Chapter 6: Permutations and Combinations

Sol. To build a 3-digit number, we need to fill the given 5 digits in 3 places: the unit’s place, the ten’s place and the hundred’s place.
(i) Repetition of the digits is allowed

Given digits are $1,2,3,4$ and 5 (all different and no zero).
Unit’s place can be filled in 5 ways. Since repetition of digits is allowed, the ten’s place can also be filled in 5




ways. For the same reason, hundred’s place can be filled in 5 ways.
∴ By the Multiplication Principle, the required number of 3-digit numbers is $5 \times 5 \times 5=125$.
(ii) Repetition of digits is not allowed

Unit’s place can be filled in 5 different ways by anyone of the given 5 digits. The ten’s place can be filled by anyone of the remaining 4 digits in 4 ways. The hundred’s place can be filled by anyone of the remaining 3 digits in 3 ways. Thus, the number of ways in which the 3 places can be filled, by the multiplication principle, is $5 \times 4 \times 3=60$. Hence, the required number of 3 -digit numbers is 60 .

Sol. To construct a 3-digit even number, we place the 6 given digits into 3 positions: units, tens, and hundreds. For even numbers, we can fill the unit’s place out of the given digits with 2 or 4 or 6 , i.e., in 3 ways. Since the digits can be repeated, the ten’s place can be filled by anyone of the 6 given digits in 6 ways. For the same reason, the hundred’s place can be filled by anyone of the 6 given digits in 6 ways.
∴ By the Multiplication Principle, the required number of 3-digit even numbers is $3 \times 6 \times 6=108$.

Sol. Since no letter can be repeated, there are 10 choices for the first letter, 9 for the second, 8 for the third, and 7 for the fourth. Applying the Fundamental Principle of Counting (Multiplication Rule), there are $10 \times 9 \times 8 \times 7=5040$ required code words in all.

Sol. Two of the ten digits (0–9) are already fixed as 6 and 7. Also no digit

$\begin{tabular}{|l|l|l|l|l|}
\hline 6 & 7 & & & \\
\hline
\end{tabular}$





appears more than once. Therefore, the remaining $10-2=8$ digits can be arranged in the three places ( $=$ Total places, i.e., $5-2$ fixed places for 6 and 7 ) after 67 in 8 , 7 and 6 ways.
Applying the Multiplication Rule (FPC), the total number of required 5-digit telephone numbers $=8 \times 7 \times 6=336$.

Sol. Each coin toss has exactly 2 possible outcomes: Head (H) or Tail (T).
We label Heads as H and Tails as T. Tossing the coin 3 times produces a tree of outcomes:

By the Multiplication Rule, when the coin is tossed three times, the total number of possible outcomes
$
=2 \times 2 \times 2=2^3=8 .
$

Note: The 8 outcomes are: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT.
(Here HHT is the outcome that the coin turns up ‘Head’ on the first toss, ‘Head’ on the second toss and ‘Tail’ on the third toss whereas HTH is the outcome that the coin turns up ‘Head’ on the first toss, ‘Tail’ on the second toss and ‘Head’ on the






third toss. Clearly, HHT and HTH are two different outcomes.) Note: For $n$ coin tosses, the total number of possible outcomes is $2^n$.

Sol. Every 2-flag signal corresponds to filling 2 positions in order from the 5 available flags. Applying the multiplication rule,

$\begin{tabular}{|l|}
\hline \\
\hline \\
\hline
\end{tabular}$ number of ways is $5 \times 4=20$.

Sol. (i) $8!=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8=40320$.
(ii) $4!-3!=(1 \times 2 \times 3 \times 4)-(1 \times 2 \times 3)$
$
=24-6=18 .
$

Sol. $3!+4!=(1 \times 2 \times 3)+(1 \times 2 \times 3 \times 4)=6+24=30$
$7!=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7=5040$
Since, $30 \neq 5040 \quad \therefore 3!+4!\neq 7!$.

Sol. $\frac{8!}{6!\times 2!}=\frac{8 \times 7 \times 6!}{6!\times 1 \times 2}=4 \times 7=28$.

Sol. Multiplying every term by $8!$ (the largest factorial in the denominators), we get
$
\begin{aligned}
& \frac{8!}{6!}+\frac{8!}{7!}=x \\
& \Rightarrow \frac{8 \times 7 \times 6!}{6!}+\frac{8 \times 7!}{7!}=x \\
& \Rightarrow \quad 56+8=x \quad \therefore \quad x=64 \text {. }
\end{aligned}
$

Sol. (i) When $n=6, r=2$, we have
$
\frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}=\frac{6 \times 5 \times 4!}{4!}=30
$
(ii) When $n=9, r=5$, we have
$
\begin{aligned}
\frac{n!}{(n-r)!} & =\frac{9!}{(9-5)!}=\frac{9!}{4!} \\
& =\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4!}{=15120}
\end{aligned}
$

Sol. The count of 3-digit numbers equals the number of ways to fill 3 positions in sequence using the 9 given digits.
This can be done in ${ }^9 \mathrm{P}_3=9 \times 8 \times 7=504$ ways.
$\therefore \quad$ The required number of 3 -digit numbers $=504$.

Sol. We have 10 digits (0 to 9) available. The number of ways of filling 4 vacant places in succession by the 10 given digits (including 0) is ${ }^{10} \mathrm{P}_4$. But these permutations will include





those also where 0 is at the thousand’s place. Such numbers are actually 3 -digit numbers (For example 0387) and must be discarded. When 0 is fixed in the thousand’s place, the remaining 9 digits can be arranged in the remaining 3 places in ${ }^9 \mathrm{P}_3$ ways.
∴ The required number of 4-digit numbers
$
\begin{aligned}
& ={ }^{10} \mathrm{P}_4-{ }^9 \mathrm{P}_3 \\
& =10 \times 9 \times 8 \times 7-9 \times 8 \times 7 \\
& =5040-504=4536
\end{aligned}
$

Alternative Method: Using place-by-place analysis with 10 digits (0–9):
The thousand place can be filled in 9 ways (excluding 0 because otherwise number will be 3 -digited only). Then hundred place can also be filled in 9 ways (including 0 and excluding the number placed in thousand place as repetition is not allowed).
Now ten’s place can be filled by any one of the remaining 8 digits in 8 ways and then unit place in 7 ways.
∴ $\quad$ The required number of 4-digit numbers $=9 \times 9 \times 8 \times 7$
$
=81 \times 8 \times 7=648 \times 7=4536
$
(By the Fundamental Principle of Counting)

Sol. For a 3-digit even number, the units place must be filled by one of the even digits: 2, 4, or 6, The remaining 5 digits can be arranged
in the remaining 2 places in ${ }^5 \mathrm{P}_2$ ways.
∴ By the Multiplication Principle, the required number of 3-digit even numbers is $3 \times{ }^5 \mathrm{P}_2=3 \times 5 \times 4=60$.

Sol. For 4-digit numbers, we arrange 5 given digits into 4 positions. This can be done in ${ }^5 \mathrm{P}_4=5 \times 4 \times 3 \times 2 =120$ ways.





For 4-digit even numbers, the unit’s place can be occupied by one of the 2 digits 2 or 4 . The remaining 4 digits can be arranged in the
remaining 3 places in ${ }^4 \mathrm{P}_3$ ways.
∴ By the Multiplication Principle, the required number of 4-digit even numbers is $2 \times{ }^4 \mathrm{P}_3=2 \times 4 \times 3 \times 2=48$.

Sol. A chairman can be chosen from 8 persons in 8 ways; then the vice-chairman is chosen from the remaining 7 in 7 ways.
∴ Using the Multiplication Principle, the selection can be made in $8 \times 7=56$ ways. OR
we can arrange 8 persons in two vacant places in ${ }^8 \mathrm{P}_2=8 \times 7=56$ ways.

Ch. V Ch.

Sol. Given, $\quad \frac{{ }^{n-1} \mathrm{P}_3}{{ }^n \mathrm{P}_4}=\frac{1}{9}$
$
\Rightarrow \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-4)}=\frac{1}{9} \Rightarrow \frac{1}{n}=\frac{1}{9}
$

Cross-multiplying gives $n = 9$.

Sol. (i) Given, $\quad{ }^5 \mathrm{P}_r=2 \times{ }^6 \mathrm{P}_{r-1}$
$
\begin{aligned}
\Rightarrow & & \frac{5!}{(5-r)!} & =2 \times \frac{6!}{\{6-(r-1)\}!} \\
\Rightarrow & & \frac{5!}{(5-r)!} & =2 \times \frac{6.5!}{(7-r)!} \\
\Rightarrow & & \frac{1}{(5-r)!} & =\frac{12}{(7-r)(6-r)(5-r)!}(\because 7-r>5-r) \\
\Rightarrow & & 1 & =\frac{12}{(7-r)(6-r)}
\end{aligned}
$




Cross-multiplying,
$
\begin{aligned}
& (7-r)(6-r)=12 \Rightarrow 42-7 r-6 r+r^2=12 \\
& \Rightarrow r^2-13 r+42=12 \quad \text { or } r^2-13 r+30=0 \\
& \Rightarrow r^2-3 r-10 r+30=0 \\
& (r-3)(r-10)=0 \\
& \therefore \quad r=3,10
\end{aligned}
$

Since ${ }^n\mathrm{P}_r$ requires $r \leq n$, $r=10$ is invalid. ∴ $r=3$.
(ii) Given, ${ }^5 \mathrm{P}_r={ }^6 \mathrm{P}_{r-1}$
$
\begin{aligned}
& \Rightarrow \quad \frac{5!}{(5-r)!}=\frac{6!}{\{6-(r-1)\}!} \\
& \Rightarrow \quad \frac{5!}{(5-r)!}=\frac{6.5!}{(7-r)!} \\
& \Rightarrow \quad \frac{1}{(5-r)!}=\frac{6}{(7-r)(6-r)(5-r)!} \\
& \Rightarrow \quad 1=\frac{6}{(7-r)(6-r)} \quad \Rightarrow \quad(7-r)(6-r)=6 \\
& \Rightarrow \quad 42-13 r+r^2=6 \quad \Rightarrow \quad r^2-13 r+36=0 \\
& \Rightarrow \quad(r-4)(r-9)=0 \quad \Rightarrow \quad r=4,9
\end{aligned}
$

Since ${ }^n\mathrm{P}_r$ requires $r \leq n$, $r=9$ is invalid. ∴ $r=4$.

Sol. EQUATION contains 8 distinct letters, which can be arranged in ${ }^8 \mathrm{P}_8$ ways.
∴ The required number of words formed
$
\begin{aligned}
& ={ }^8 \mathrm{P}_8=8!=1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \\
& =40320
\end{aligned}
$

Sol. MONDAY has 6 different letters.
(i) 4 letters out of 6 can be arranged in ${ }^6 \mathrm{P}_4$ ways.
$\therefore \quad$ The required number of words $={ }^6 \mathrm{P}_4$
$
=6 \times 5 \times 4 \times 3=360 .
$
(ii) 6 letters can be arranged among themselves in ${ }^6 \mathrm{P}_6$ ways.
∴ The required number of words $={ }^6 \mathrm{P}_6=6$ !
$
=1 \times 2 \times 3 \times 4 \times 5 \times 6=720 \text {. }
$
(iii) The first place can be filled by anyone of the two vowels O or A in 2 ways. The remaining 5 letters can be arranged in the remaining 5 places II to VI in ${ }^5 \mathrm{P}_5$
$
=5!\text { ways. }
$
∴ By the Multiplication Principle, the required number of words $=2 \times 5!=2 \times 1 \times 2 \times 3 \times 4 \times 5=240$.

Sol. MISSISSIPPI has 11 letters with repetitions: I occurs 4 times, S occurs 4 times, P occurs twice, M occurs once.
$
\begin{gathered}
\therefore \text { Total number of permutations }=\frac{11!}{4!4!2!} \left\lvert\, \frac{n!}{\mathrm{p}!q!\mathrm{r}!}\right. \\
=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1 \times 2 \times 1} \\
=34650
\end{gathered}
$

Next, we count the arrangements where all four I’s are grouped together. Treating the four I’s as one letter (Units).





(IIII) SSSS P P M
These are $1+4+2+1=8$ letters (Units)
We have 8 letters (Units) which can be arranged in $\frac{8!}{4!2!}$ ways.
The four I’s can be arranged among themselves in only one way $\left(\because \frac{4!}{4!}=1\right)$
∴ The number of permutations in which the four I’s come together
$
=\frac{8!}{4!2!}=\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 2 \times 1}=840
$

Hence, the number of arrangements where the four I’s are NOT together equals the total arrangements minus the ‘together’ case:
$
=34650-840=33810 .
$

Sol. PERMUTATIONS has 12 letters, with T occurring twice.
(i) With P fixed at the start and S at the end, the remaining 10 letters (T appearing twice) are arranged in the 10 middle positions. This can be done in $\frac{10!}{2!}$ ways.

$\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|l|l|}
\hline P & & & & & & & & & & & S \\
\hline
\end{tabular}$

$
\begin{aligned}
& \therefore \text { The required number of words }=\frac{10!}{2!} \\
& \quad=\frac{1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10}{1 \times 2} \\
& \quad=1814400
\end{aligned}
$




(ii) PERMUTATIONS has 5 distinct vowels (E, U, A, I, O) and 7 consonants (T appears twice). Treating all 5 vowels as one block (EUAIO),

$\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline EUAIO & P & R & M & T & T & N & S \\
\hline
\end{tabular}$

this block + 7 consonants = 8 objects, with T occurring twice. These 8 objects can be arranged in $\frac{8!}{2!}$ ways. Corresponding to each of these arrangements, the 5 vowels, all distinct, can be arranged in ${ }^5 \mathrm{P}_5=5$ ! ways. Therefore, by multiplication principle, the required number of words
$
\begin{aligned}
& =\frac{8!}{2!} \times 5! \\
& =\frac{(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8)(1 \times 2 \times 3 \times 4 \times 5)}{1 \times 2} \\
& =20160 \times 120=2419200
\end{aligned}
$
(iii) With exactly 4 letters between P and S, therefore, P and S can occupy 1st and 6th places or 2nd and 7th places
or 3rd and 8th places
or 4th and 9th places or 5th and 10th places or 6th and 11th places or 7th and 12th places.
This gives 7 possible position pairs for P and S. P and S can also swap their positions in 2 ways. The remaining 10 positions are filled by the other 10 letters (T twice) in $\frac{10!}{2!}$ ways.
$\therefore \quad$ The required number of words $=7 \times 2 \times \frac{10!}{2!}$




$
\begin{aligned}
&\text { Class } 11\\
&\text { Chapter } 7 \text { – Permutations and Combinations }\\
&\begin{aligned}
& =7 \times(1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10) \\
& =7 \times 720 \times 56 \times 90=25401600
\end{aligned}\\
&\text { }\\
&14
\end{aligned}
$

Sol. ${ }^n \mathrm{C}_8={ }^n \mathrm{C}_2$
⇒ Either $8=2$ which is false
$
\left[\because \quad{ }^n \mathrm{C}_p={ }^n \mathrm{C}_q \Rightarrow p=q \text { or } p+q=n\right]
$
or $\quad 8+2=n \quad \therefore n=10$
$
\therefore \quad{ }^n \mathrm{C}_2={ }^{10} \mathrm{C}_2=\frac{10 \times 9}{2 \times 1}=45 .
$

Sol. (i) $\quad{ }^{2 n} \mathrm{C}_3:{ }^n \mathrm{C}_3=12: 1$
$
\begin{array}{lrl}
\Rightarrow & \frac{2 n(2 n-1)(2 n-2)}{3 \times 2 \times 1} & : \frac{n(n-1)(n-2)}{3 \times 2 \times 1}=12: 1 \\
\Rightarrow & & \frac{4 n(2 n-1)(n-1)}{n(n-1)(n-2)}=\frac{12}{1} \\
\Rightarrow & \frac{4(2 n-1)}{n-2}=12 \\
\Rightarrow & \frac{2 n-1}{n-2}=3 \Rightarrow & 2 n-1=3(n-2) \\
\Rightarrow & & 2 n-1=3 n-6 \\
\therefore & & n=5 .
\end{array}
$




(ii) ${ }^{2 n} \mathrm{C}_3:{ }^n \mathrm{C}_3=11: 1$
$
\Rightarrow \quad \frac{2 n(2 n-1)(2 n-2)}{3 \times 2 \times 1}: \frac{n(n-1)(n-2)}{3 \times 2 \times 1}=11: 1
$

Factoring out 2 from $(2n-2)$,
$
\begin{array}{lllc}
\Rightarrow & & \frac{4 n(2 n-1)(n-1)}{n(n-1)(n-2)}=11 \\
& & \frac{4(2 n-1)}{n-2}=11 & \Rightarrow \\
& & \Rightarrow & 4(2 n-1)=11(n-2) \\
\Rightarrow & -3 n=-18 & \Rightarrow & 8 n-4=11 n-22 \\
\therefore & & & 3 n=18 \\
& & n=6
\end{array}
$

Sol. Since any two points on a circle determine a unique chord:
Now, a chord can be drawn by joining any two of the 21 given points.
∴ $\quad$ The required number of chords $={ }^{21} \mathrm{C}_2$
[ ∵ Every two points can be joined to form a straight line]
$
=\frac{21 \times 20}{2 \times 1}=210
$

Sol. We select 3 boys from 5 in ${ }^5 \mathrm{C}_3$ ways and 3 girls can be selected out of 4 girls in ${ }^4 \mathrm{C}_3$ ways.
∴ Applying the Multiplication Principle, the required number of ways of selecting a team $={ }^5 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3$
$
\begin{aligned}
& ={ }^5 \mathrm{C}_2 \times{ }^4 \mathrm{C}_1 \quad\left[\because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right] \\
& =\frac{5 \times 4}{2 \times 1} \times 4=40
\end{aligned}
$

Sol. We select 3 red balls from 6 in ${ }^6 \mathrm{C}_3$ ways.
3 white balls out of 5 can be selected in ${ }^5 \mathrm{C}_3$ ways.
3 blue balls out of 5 can be selected in ${ }^5 \mathrm{C}_3$ ways.






∴ Applying the Multiplication Principle, 9 balls can be selected in
$
\begin{aligned}
{ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3={ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2 \quad & {\left[\because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right] } \\
& =\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{5 \times 4}{2 \times 1} \times \frac{5 \times 4}{2 \times 1} \\
& =20 \times 10 \times 10=2000 \text { ways. }
\end{aligned}
$

Sol. A standard deck has 4 aces and 48 non-ace cards. One ace can be selected out of 4 in ${ }^4 \mathrm{C}_1$ ways and $5-1=4$ other cards out of 48 in ${ }^{48} \mathrm{C}_4$ ways.
∴ Applying the Multiplication Principle, one ace and 4 other cards can be selected in ${ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_4$
$
=4 \times \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}=778320 \text { ways. }
$

Sol. Among 17 players, 5 are bowlers and 12 are non-bowlers. 4 bowlers out of 5 can be selected in ${ }^5 \mathrm{C}_4$ ways and $11-4=7$ others players can be selected out of 12 other players in ${ }^{12} \mathrm{C}_7$ ways.
∴ Applying the Multiplication Principle, a cricket team of 11 can be selected in ${ }^5 \mathrm{C}_4 \times{ }^{12} \mathrm{C}_7$
$
\begin{aligned}
& ={ }^5 \mathrm{C}_1 \times{ }^{12} \mathrm{C}_5 \quad\left[\because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right] \\
& =5 \times \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}=3960
\end{aligned}
$

Sol. We choose 2 black balls from 5 in ${ }^5 \mathrm{C}_2$ ways and 3 red balls out of 6 can be selected in ${ }^6 \mathrm{C}_3$ ways.
∴ Applying the Multiplication Principle, the required number of selections is ${ }^5 \mathrm{C}_2 \times{ }^6 \mathrm{C}_3$
$
=\frac{5 \times 4}{2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}=200 .
$

Sol. With 2 courses already compulsory, the student must pick 3 more from the remaining 7. This he can do in ${ }^7 \mathrm{C}_3=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35$ ways.

Sol. DAUGHTER has 3 vowels (A, U, E) and 5 consonants (D, G, H, T, R).
2 vowels out of 3 can be selected in ${ }^3 \mathrm{C}_2={ }^3 \mathrm{C}_1=3$ ways. 3 consonants out of 5 can be selected in ${ }^5 \mathrm{C}_3={ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1} =10$ ways.
∴ The number of selections of 2 vowels and 3 consonants is $3 \times 10=30$.
Each of the 30 combinations has 5 letters that can be arranged in ${ }^5 \mathrm{P}_5=5!=1 \times 2 \times 3 \times 4 \times 5 =120$ ways.
∴ The required number of different words is
$
30 \times 120=3600
$

Note: In this problem, we first chose the letters (combinations) and then arranged them (permutations).

Sol. EQUATION has 8 distinct letters: 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). We treat the 5 vowels (EUAIO) as one block and the 3 consonants (QTN) as another block. These 2 blocks can be arranged in $2! = 2$ ways. Within each arrangement, the 5 vowels can permute among themselves in $5!$ ways and the 3 consonants in $3!$ ways.
∴ Applying the Multiplication Principle, the required number of words is $2!\times 5!\times 3!=2 \times 120 \times 6=1440$.

Sol. (i) We can choose 3 girls from 4 in ${ }^4 \mathrm{C}_3$ ways and 7-3 $=4$ boys can be selected out of 9 in ${ }^9 \mathrm{C}_4$ ways. Therefore, the required number of ways is
$
\begin{aligned}
{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4 & ={ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_4 \quad\left[\because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right] \\
& =4 \times \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}=504
\end{aligned}
$
(ii) “At least 3 girls” means the committee has either 3 or 4 girls:
(a) 3 girls and 4 boys (b) 4 girls and 3 boys

3 girls and 4 boys can be selected in ${ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4$ ways.
4 girls and 3 boys can be selected in ${ }^4 \mathrm{C}_4 \times{ }^9 \mathrm{C}_3$ ways.
∴ The required number of ways
$
\begin{aligned}
& ={ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_4 \times{ }^9 \mathrm{C}_3 \\
& ={ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_0 \times{ }^9 \mathrm{C}_3 \\
& =4 \times \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}+1 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \\
& =504+84=588
\end{aligned}
$
(iii) “At most 3 girls” means 0, 1, 2, or 3 girls. The valid cases are:
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) no girl and 7 boys






3 girls and 4 boys can be selected in ${ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4$ ways.
2 girls and 5 boys can be selected in ${ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_5$ ways.
1 girl and 6 boys can be selected in ${ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_6$ ways.
No girl and 7 boys can be selected in ${ }^4 \mathrm{C}_0 \times{ }^9 \mathrm{C}_7$ ways.
∴ The required number of ways
$
\begin{aligned}
= & { }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_5+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_6+{ }^4 \mathrm{C}_0 \times{ }^9 \mathrm{C}_7 \\
= & { }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_3+{ }^4 \mathrm{C}_0 \times{ }^9 \mathrm{C}_2 \\
\quad\left[\because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right] & \\
= & 4 \times \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1}+\frac{4 \times 3}{2 \times 1} \times \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \\
& +4 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1}+1 \times \frac{9 \times 8}{2 \times 1} \\
= & 504+756+336+36=1632 .
\end{aligned}
$

Sol. Arranging the letters of EXAMINATION alphabetically: AAEIIMNNOTX
Total number of letters in this word is 11 .
∴ The count of words before the first E-word equals the count of words starting with A (since A is the only letter before E here). With A in position 1, the remaining 10 letters (two I’s and two N’s) are arranged in the other 10 spots.

A $\times \times \times \times \times \times \times \times \times \times \times$
∴ The required number of words
$
\begin{aligned}
& =\frac{10!}{2!2!} \\
& =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!}{2 \times 1 \times 2!} \\
& =907200
\end{aligned}
$

Sol. Numbers divisible by 10 must have ‘ 0 ‘ in the unit’s place.
$
\times \times \times \times \times 0
$

The remaining 5 digits fill the 5 remaining positions in ${ }^5\mathrm{P}_5 = 5!$ ways.
$\therefore \quad$ The required number of 6 -digit numbers $=5!=120$.

Sol. We select 2 vowels from 5 in
$
{ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1}=10 \text { ways. }
$

2 consonants out of 21 can be selected in
$
{ }^{21} \mathrm{C}_2=\frac{21 \times 20}{2 \times 1}=210 \text { ways. }
$
∴ The number of selections of 2 vowels and 2 consonants is $10 \times 210=2100$.

Each of the 2100 combinations contains 4 letters that can be arranged in
$
{ }^4 \mathrm{P}_{4=} 4!=1 \times 2 \times 3 \times 4=24 \text { ways. }
$

Therefore, the required number of different words $=2100 \times 24=50400$.

Sol. The possible ways to pick 8 questions (with at least 3 from each part) are:





\begin{tabular}{rcc}
& Part I & Part II \\
(i) & 5 & 7 \\
(ii) & 3 & 5 \\
(iii) & 4 & 4 \\
5 & 3
\end{tabular}
( ∵ A student has to select at least 3 questions from each part.)
The required number of ways
$
\begin{aligned}
& ={ }^5 \mathrm{C}_3 \times{ }^7 \mathrm{C}_5+{ }^5 \mathrm{C}_4 \times{ }^7 \mathrm{C}_4+{ }^5 \mathrm{C}_5 \times{ }^7 \mathrm{C}_3 \\
& ={ }^5 \mathrm{C}_2 \times{ }^7 \mathrm{C}_2+{ }^5 \mathrm{C}_1 \times{ }^7 \mathrm{C}_3+{ }^5 \mathrm{C}_0 \times{ }^7 \mathrm{C}_3 \\
& \quad \mid \because{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r} \\
& =\frac{5 \times 4}{2 \times 1} \times \frac{7 \times 6}{2 \times 1}+5 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}+1 \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1} \\
& =10 \times 21+5 \times 35+1 \times 35 \\
& =210+175+35=420
\end{aligned}
$

Sol. A standard deck has 4 kings and 48 non-king cards. One king can be selected out of 4 in ${ }^4 \mathrm{C}_1$ ways and $5-1=4$ other cards out of 48 in ${ }^{48} \mathrm{C}_4$ ways.
∴ Applying the Multiplication Principle, one king and 4 other cards can be selected in ${ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_4=4 \times 194580=778320$ ways.

Sol. In a row of 9 seats, positions 2, 4, 6, and 8 are the even-numbered places. Four women can be arranged in four even places in ${ }^4 \mathrm{P}_4$ ways. Five men can be arranged in the remaining five odd places in ${ }^5 \mathrm{P}_5$ ways. Using the Fundamental Counting Principle (Multiplication), the required number of arrangements is
$
\begin{aligned}
{ }^4 \mathrm{P}_4 \times{ }^5 \mathrm{P}_5=4!\times 5! & =(4 \times 3 \times 2 \times 1) \times(5 \times 4 \times 3 \times 2 \times 1) \\
& =24 \times 120=2880
\end{aligned}
$

Sol. There are exactly two scenarios based on the given condition:
(i) the particular 3 students join
(ii) the particular 3 students do not join.

Case 1 (all 3 join): We need to pick $7 = 10-3$ more students from the remaining $22 = 25-3$. This can be done in ${ }^{22} \mathrm{C}_7$ ways.
Case 2 (none join): We pick all 10 students from the remaining $22 = 25-3$. This can be done in ${ }^{22} \mathrm{C}_{10}$ ways.
∴ $\quad$ The required number of ways $={ }^{22} \mathrm{C}_7+{ }^{22} \mathrm{C}_{10}$
[Addition Principle]
$
\begin{aligned}
& =\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\
& +\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \\
& =170544+646646=817190
\end{aligned}
$

Sol. ASSASSINATION has 13 letters: A×3, S×4, I×2, N×2, and one each of T and O. Since all 4 S’s must stay together, we treat (SSSS) as a single unit. This unit combined with the other 9 letters gives 10 objects — (SSSS), A, A, A, I, I, N, N, T, O — which can be arranged in $\frac{10!}{3!2!2!}$ ways.
The 4 S’s themselves can only be arranged in 1 way
$\therefore \quad$ The required number of ways $=\frac{10!}{3!2!2!} \times 1$
$
\begin{aligned}
& =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1) \times(2 \times 1) \times(2 \times 1)} \\
& =151200
\end{aligned}
$