Exercise 4.3 - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$
\text { Ex } 4.3
$
Question 1.
$
\text { If }{ }^{\mathrm{n}} \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_9 \text { find }{ }^{21} \mathrm{C}_{\mathrm{n}}
$
Solution:
$
\begin{aligned}
& { }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{y}} \Rightarrow \mathrm{x}=\mathrm{y} \text { or } \mathrm{x}+\mathrm{y}=\mathrm{n} \\
& \text { Here }{ }^{\mathrm{n}} \mathrm{C}_{12}={ }^{\mathrm{n}} \mathrm{C}_9 \Rightarrow 12 \neq 9 \text { so } 12+9=\mathrm{n} \text { (i.e) } \mathrm{n}=21 \\
& \therefore{ }^{21} \mathrm{C}_n={ }^{21} \mathrm{C}_{21}=1 \quad\left[{ }^n \mathrm{C}_{\boldsymbol{n}}=\mathbf{1}\right]
\end{aligned}
$
Question 2.
If ${ }^{15} \mathrm{C}_{2 \mathrm{r}-1}={ }^{15} \mathrm{C}_{2 \mathrm{r}+4}$, find $\mathrm{r}$.
Solution:
$
\begin{aligned}
& { }^n \mathrm{C}_x={ }^n \mathrm{C}_y \Rightarrow x=y \text { or } x+y=n \\
& \text { Here }{ }^{15} \mathrm{C}_{2 r-1}={ }^{15} \mathrm{C}_{2 r+4} \\
& \Rightarrow \quad \begin{array}{l}
2 r-1+2 r+4=15 \\
4 r+3=15
\end{array} \\
& \begin{array}{l}
4 r+3=15 \\
4 r=15-3=12 \Rightarrow r=\frac{12}{4}=3
\end{array} \\
& r=3 \\
&
\end{aligned}
$
Question 3.
$
\text { If } \mathrm{P}_{\mathrm{r}}=720 \text { and }{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=120 \text {, find } \mathrm{n}, \mathrm{r} \text {. }
$
Solution:
$
\begin{aligned}
& { }^n \mathrm{P}_r=\frac{n !}{(n-r) !} ;{ }^n \mathrm{C}_r=\frac{n !}{r !(n-r) !} \\
& \Rightarrow \frac{{ }^n \mathrm{P}_r}{r !}={ }^n C_r \\
& \text { (i.e) } \frac{720}{r !}=120 \Rightarrow \frac{720}{120}=r ! \\
& r !=6=3 ! \Rightarrow r=3 \\
& n \mathrm{P}_3=720 \\
& n(n-1)(n-2)=720=10 \times 9 \times 8 \Rightarrow n=10
\end{aligned}
$

Question 4.
Prove that ${ }^{15} \mathrm{C}_3+2 \times{ }^{15} \mathrm{C}_4+{ }^{15} \mathrm{C}_5={ }^{17} \mathrm{C}_5$
Solution:
$
\begin{aligned}
\text { LHS } & =\frac{15 \times 14 \times 13}{3 \times 2 \times 1}+2 \times \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}+\frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \\
& =\frac{15 \times 14 \times 13}{1 \times 2 \times 3}\left[1+\frac{24}{4}+\frac{132}{20}\right] \\
& =\frac{15 \times 14 \times 13}{1 \times 2 \times 3}\left[\frac{20+120+132}{20}\right] \\
& =\frac{15 \times 14 \times 13 \times 272}{1 \times 2 \times 3 \times 4 \times 5}=\frac{17 \times 16 \times 15 \times 14 \times 13}{5 !}={ }^{17} \mathrm{C}_5=\text { RHS }
\end{aligned}
$
Question 5.
Prove that ${ }^{35} C_5+\sum_{r=0}^4{ }^{(39-r)} C_4={ }^{40} C_5$.
Solution:
$
\text { LHS }={ }^{35} \mathrm{C}_5+{ }^{39} \mathrm{C}_4+{ }^{38} \mathrm{C}_4+{ }^{37} \mathrm{C}_4+{ }^{36} \mathrm{C}_4+{ }^{35} \mathrm{C}_4
$
Now we know ${ }^n \mathrm{C}_r+{ }^n \mathrm{C}_{r-1}={ }^{n+1} \mathrm{C}_r$
So ${ }^{35} \mathrm{C}_5+{ }^{35} \mathrm{C}_4={ }^{36} \mathrm{C}_5$
Now ${ }^{36} \mathrm{C}_5+{ }^{36} \mathrm{C}_4={ }^{37} \mathrm{C}_5$
${ }^{37} \mathrm{C}_5+{ }^{37} \mathrm{C}_4={ }^{38} \mathrm{C}_5$
${ }^{38} \mathrm{C}_5+{ }^{38} \mathrm{C}_4={ }^{39} \mathrm{C}$
${ }^{39} \mathrm{C}_5+{ }^{39} \mathrm{C}_4={ }^{40} \mathrm{C}_5=$ RHS
Question 6.
If ${ }^{(n+1)} C_8:{ }^{(n-3)} P_4=57: 16$, find the value of $n$.
Solution:

$\begin{aligned}
& \text { Given }{ }^{(n+1)} \mathrm{C}_8:\left({ }^{(n-3)} \mathrm{P}_4=57: 16\right. \\
& \Rightarrow \quad \frac{{ }^{n+1} \mathrm{C}_8}{{ }^{n-3} \mathrm{P}_4}=\frac{57}{16} \\
& \frac{(n+1) !}{8 !(n+1-8) !} / \frac{(n-3) !}{(n-3-4) !}=\frac{57}{16} \\
& \frac{(n+1) !}{8 !(n-7) !} \times \frac{(n-7) !}{(n-3) !}=\frac{57}{16} \\
& \frac{(n+1) !}{(n-3) !}=8 ! \frac{57}{16} \\
& \frac{(n+1)(n)(n-1)(n-2)(n-3) !}{(n-3) !}=8 ! \frac{57}{16} \\
& (n+1)(n)(n-1)(n-2)=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 3 \times 19}{16} \\
& =18 \times 21 \times 20 \times 19 \\
& =21 \times 20 \times 19 \times 18 \\
& \Rightarrow n=20 \\
&
\end{aligned}$

Question 7.
Prove that ${ }^{2 n} C_n=\frac{2^n \times 1 \times 3 \ldots(2 n-1)}{n !}$

Solution:
$
\begin{aligned}
& \text { LHS }={ }^{2 n} \mathrm{C}_n=\frac{2 n !}{n !(2 n-n) !}=\frac{2 n !}{n ! n !} \\
& =\frac{(2 n)(2 n-1)(2 n-2)(2 n-3) \ldots .4 .3 .2 .1}{n ! n !}
\end{aligned}
$
Numerator has $2 n$ terms in which $n$ terms are even and $n$ terms are odd. Taking one 2 from the $n$ even terms we get
$
\begin{aligned}
& =\frac{2(n)(2 n-1)(2)(n-1)(2 n-3) \ldots 2(2) \cdot 3.2(2) .1}{n ! n !} \\
& =\frac{2^n[(n)(n-1)(n-2) \ldots 2.1][(2 n-1)(2 n-3) \ldots 3.1]}{n ! n !} \\
& =\frac{2^n \times n !(2 n-1)(2 n-3) \ldots 3.1}{n ! n !} \\
& =\frac{2^n \times 1 \times 3 \times 5 \ldots(2 n-3)(2 n-1)}{n !}=\text { RHS }
\end{aligned}
$
Question 8. Prove that if $1 \leq \mathrm{r} \leq \mathrm{n}$
then $n \times \quad(n-1) C_{r-1}=(n-r+1) C_{r-1}$.
Solution:
To Prove $n\left[{ }^{n-1} \mathrm{C}_{r-1}\right]=(n-r+1)\left[{ }^n \mathrm{C}_{r-1}\right]$
LHS: $n\left[\frac{(n-1) !}{(r-1) !(n-1-(r-1) !(n-1-r+1)}\right]$
$
\begin{gathered}
\quad=\frac{n(n-1) !}{(r-1) !(n-r) !}=\frac{n !}{(r-1) !(n-r) !} \\
\text { RHS: } \left.(n-r+1)\left[{ }^n \mathrm{C}_{r-1}\right)\right] \quad=(n-r+1)\left[\frac{n)}{(r-1) !(n-r-1) !(n-r+1)}\right]
\end{gathered}
$

$\begin{aligned}
& =(n-r+1)\left[\frac{n !}{(r-1) !(n-r+1) !}\right] \\
& =\frac{(n-r+1) n !}{(r-1) !(n-r+1)(n-r) !} \\
& =\frac{n !}{(r-1) !(n-r) !} \\
(1) & =(2) \Rightarrow \text { LHS }=\text { RHS }
\end{aligned}$

Question 9.
(i) A Kabaddi coach has 14 players ready to play. How many different teams of 7 players could the coach put on the court?
Solution:
No. of players in the team $=14$
We need 7 players
So selecting 7 from 14 players can be done is ${ }^{14} C_7=3432$ ways
(ii) There are 15 persons in a party and if, each 2 of them shakes hands with each other, how many handshakes happen in the party?
Solution:
Total No. of persons $=15$
Each two persons shake hands
$\therefore$ No. of hand shakes $=\frac{15}{2}=\frac{15 \times 14}{2 \times 1}=105$
(iii) How many chords can be drawn through 20 points on a circle?
Solution:
A chord is a line join of 2 points
No. of points given $=20$
Selecting 2 from 20 can be done in ${ }^{20} \mathrm{C}_2$ ways
So number of chords $={ }^{20} \mathrm{C}_2=\frac{20 \times 19}{2 \times 1}=190$
(iv) In a parking lot one hundred, one year old cars are parked. Out of them five are to be chosen at random for to check its pollution devices. How many different set of five cars are possible?
Solution:
Number of cars $=100$
Select 5 from 100 cars can be done in ${ }^{100} C_5$ ways
(v) How many ways can a team of 3 boys, 2 girls and 1 transgender be selected from 5 boys, 4 girls and 2 transgenders?
Solution:
We have 5 boys, 4 girls and 2 transgenders. We need 3 boys, 2 girls and 1 transgender The selection can be done as follows Selecting 3 boys from 5 boys can be done is ${ }^5 \mathrm{C}_3$ ways
$
{ }^5 C_3={ }^5 C_2=\frac{5 \times 4}{2 \times 1}=10
$
Selecting 2 girls from 4 girls can be done in ${ }^4 C_2$ ways
$
{ }^4 C_2=\frac{4 \times 3}{2 \times 1}=6
$
Selecting 1 transgender from 2 can be done in ${ }^2 \mathrm{C}_1=2$ ways
$\therefore$ Selecting 3 boys, 2 girls and 1 transgender can be done in $10 \times 6 \times 2=120$ ways
Question 10.
Find the total number of subsets of a set with
(i) 4 elements
(ii) 5 elements
(iii) $n$ elements
Hint. ${ }^n \mathrm{C}_0+{ }^n \mathrm{C}_1+{ }^n \mathrm{C}_2+\ldots .{ }^n \mathrm{C}_n=2 n$
Solution:
If a set has $n$ elements then the number of its subsets $=2^{\mathrm{n}}$
(i) Here $n=4$
So number of subsets $=2^4=16$
(ii) $\mathrm{n}=5$
So number of subsets $=2^5=32$
(iii) $\mathrm{n}=\mathrm{n}$
So number of subsets $=2$
Question 11.
A trust has 25 members.
(i) How many ways 3 officers can be selected?
Solution:
Selecting 3 from 25 can be done in ${ }^{25} \mathrm{C}_3$ ways
$
{ }^{25} \mathrm{C}_3=\frac{25 \times 24 \times 23}{3 \times 2 \times 1}=2300
$
(ii) In how many ways can a President, Vice President and a secretary be selected?
Solution:
From the 25 members a president can be selected in 25 ways
After the president is selected, 24 persons are left out.

So a Vice President can be selected in (from 24 persons) 24 ways.
After the selection of Vice President 23 persons are left out
So a secretary can be selected (from the remaining 23 persons) in 23 ways
So a president, Vice president and a secretary can be selected in ${ }^{25} \mathrm{P}_3$ ways ${ }^{25} \mathrm{P}_3=25 \times 24 \times 23=13800$ ways
Question 12.
How many ways a committee of six persons from 10 persons can be chosen along with a chair person and a secretary?
Solution:
Selecting a chair person from the 10 persons can be done in 10 ways
After the selection of chair person only 9 persons are left out so selecting a secretary (from the remaining a persons) can be done in 9 ways.
The remaining persons $=8$
Totally we need to select 6 persons
We have selected 2 persons.
So we have to select 4 persons
Selecting 4 from 8 can be done in ${ }^8 \mathrm{C}_4$ ways
$
\begin{aligned}
\therefore \text { Total number of selection } & =10 \times 9 \times{ }^8 \mathrm{C}_4 \\
& =\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=6300
\end{aligned}
$
Question 13.
How many different selections of 5 books can be made from 12 different books if,
Solution:
No. of books given $=12$
No. of books to be selected $=5$
(i) Two particular books are always selected?
Solution:
So we need to select 3 more books from (12-2) 10 books which can be done in ${ }^{10} \mathrm{C}_3$ ways
$
{ }^{10} \mathrm{C}_3=\frac{10 \times 9 \times 8}{3 \times 2 \times 1}=120
$

(ii) Two particular books are never selected?
Solution:
Two particular books never to be selected.
So only 10 books are there and we have to select 5 books which can be done in ${ }^{10} \mathrm{C}_5$ ways
$
{ }^{10} \mathrm{C}_5=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=252
$
Question 14.
There are 5 teachers and 20 students. Out of them a committee of 2 teachers and 3 students is to be formed. Find the number of ways in which this can be done. Further find in how many of these committees
(i) a particular teacher is included?
(ii) a particular student is excluded?
Solution:
No. of teachers $=5$
No of students $=20$
We need to select 2 teachers and 3 students
Selecting 2 from 5 teachers can be done in ${ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1}=10$ ways
Selecting 3 from 20 students can be done in ${ }^{20} \mathrm{C}_3=\frac{20 \times 19 \times 18}{3 \times 2 \times 1}=1140$ ways
So 2 teachers and 3 students together can be selected in ${ }^5 C_2 \times{ }^{20} C_3$ ways
$
=10 \times 1140=11400 \text { ways }
$
(i) A particular teacher should be included. So from the remaining 4 teachers one teacher is to be selected which can be done in ${ }^4 C_1=4$ ways
Selecting 3 students from 20 can be done in ${ }^{20} \mathrm{C}_3=\frac{20 \times 19 \times 18}{3 \times 2 \times 1} \quad 1140$ ways
So selecting 2 teachers and 3 students can be done in $4 \times 1140=4560$ ways

(ii) particular student should be excluded.
So we have to select 3 students from 19 students which can be done in ${ }^{19} \mathrm{C}_3$ ways
$
{ }^{19} \mathrm{C}_3=\frac{19 \times 18 \times 17}{3 \times 2 \times 1}=969 \text { ways }
$
Two teachers from 5 can be selected in ${ }^5 C_2=\frac{5 \times 4}{2 \times 1}=10$ ways
$\therefore 2$ teachers and 3 students can be selected in $969 \times 10=9690$ ways
Question 15 .
In an examination a student has to answer 5 questions, out of 9 questions in which 2 are compulsory. In how many ways a students can answer the questions?
Solution:
No. of questions given $=9$
No. of questions to be answered $=5$
But 2 questions are compulsory
So the student has to answer the remaining 3 questions $(5-2=3)$ from the remaining $7(9-2=7)$ questions which can be done in ${ }^7 C_3$ ways
$
{ }^7 C_3=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways }
$
Question 16.
Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly three aces in each combination.
Solution:
No. of cards $=52:$ In that number of aces $=4$
No. of cards needed $=5$
In that 5 cards number of aces needed $=3$

So the 3 aces can be selected from 4 aces in ${ }^4 \mathrm{C}_3={ }^4 \mathrm{C}_1=4$ ways
So the remaining $=5-3=2$
This 2 cards can be selected in ${ }^{48} \mathrm{C}_2$ ways
${ }^{48} \mathrm{C}_2=\frac{48 \times 47}{2 \times 1}=1128$ ways
$\therefore$ No. of ways in which the 5 cards can be selected $=\left({ }^{48} \mathrm{C}_2\right)\left({ }^4 \mathrm{C}_3\right)$
$
=\frac{48 \times 47}{2 \times 1} \times 4=4512
$
Question 17.
Find the number of ways of forming a committee of 5 members out of 7 Indians and 5 Americans, so that always Indians will be the majority' in the committee.
Solution:
We need a majority of Indian's which is obtained as follows.

The possible ways are (5I) or (4I and $1 \mathrm{~A}$ ) or ( $3 \mathrm{I}$ and $2 \mathrm{~A}$ )
$
\begin{gathered}
=\left({ }^7 \mathrm{C}_5\right)\left({ }^5 \mathrm{C}_0\right)+\left({ }^7 \mathrm{C}_4\right)\left({ }^5 \mathrm{C}_1\right)+\left({ }^7 \mathrm{C}_3\right)\left({ }^5 \mathrm{C}_2\right) \\
{ }^7 \mathrm{C}_5={ }^7 \mathrm{C}_2=\frac{7 \times 6}{2 \times 1}=21 ;{ }^7 \mathrm{C}_3={ }^7 \mathrm{C}_4=\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \\
{ }^5 \mathrm{C}_0=1 ;{ }^5 \mathrm{C}_1=5 ;{ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1}=1 \mathrm{C} \\
\therefore \text { The possible ways (21) (1) }+(35)(5)+(35)(10)=21+175+350=546
\end{gathered}
$
Question 18 .
A committee of 7 peoples has to be formed from 8 men and 4 women. In how many ways can this be done when the committee consists of
(i) exactly 3 women?
(ii) at least 3 women?
(iii) at most 3 women?
Solution:

(i)

We need a committee of 7 people with 3 women and 4 men.
This can be done in $\left({ }^4 \mathrm{C}_3\right)\left({ }^8 \mathrm{C}_4\right)$ ways
$
\begin{aligned}
{ }^4 C_3 & ={ }^4 C_1=4 \\
{ }^8 C_4 & =\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \quad 70
\end{aligned}
$
The number of ways $=(70)(4)=280$
(ii) Atleast 3 women

So the possible ways are ( $3 \mathrm{~W}$ and $4 \mathrm{M}$ ) or ( $4 \mathrm{~W}$ and $3 \mathrm{M}$ )
$
\begin{aligned}
& \text { (i.e) }\left({ }^4 C_3\right)\left({ }^8 C_4\right)+\left({ }^4 C_4\right)\left({ }^8 C_3\right) \\
& { }^4 C_3={ }^4 C_1=4 ;{ }^4 C_4=1 \\
& { }^8 C_4=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=70 \\
& { }^8 C_3=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56
\end{aligned}
$
The number of ways (4) (70) + (1) (56) $=280+56=336$
(iii) Atmost 3 women

The possible ways are $(0 \mathrm{~W} 8 \mathrm{M})$ or $(1 \mathrm{~W} 6 \mathrm{M})$ or $(2 \mathrm{~W} 5 \mathrm{M})$ or $(3 \mathrm{~W} 4 \mathrm{M})$
$
\begin{aligned}
&=\left(\begin{array}{l}
4 \\
0
\end{array}\right)\left(\begin{array}{l}
8 \\
7
\end{array}\right)+\left(\begin{array}{l}
4 \\
1
\end{array}\right)\left(\begin{array}{l}
8 \\
6
\end{array}\right)+\left(\begin{array}{l}
4 \\
2
\end{array}\right)\left(\begin{array}{l}
8 \\
5
\end{array}\right)+\left(\begin{array}{l}
4 \\
3
\end{array}\right)\left(\begin{array}{l}
8 \\
4
\end{array}\right) \\
&=\left[\left(\begin{array}{l}
n \\
r
\end{array}\right)={ }^n \mathrm{C}_r\right] \\
&{ }^4 \mathrm{C}_0=1 ;{ }^4 \mathrm{C}_1={ }^4 \mathrm{C}_3=4 ;{ }^4 \mathrm{C}_2=\frac{4 \times 3}{2 \times 1}=6 \\
&{ }^8 \mathrm{C}_7={ }^8 \mathrm{C}_1=8 ;{ }^8 \mathrm{C}_6={ }^8 \mathrm{C}_2=\frac{8 \times 7}{2 \times 1}=28 \\
&{ }^8 \mathrm{C}_5={ }^8 \mathrm{C}_3=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56 \\
&{ }^8 \mathrm{C}_4=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=70
\end{aligned}
$
$\therefore$ The possible ways are
(1) (8) $+(4)(28)+(6)(56)+(4)(70)=8+112+336+280=736$ ways
Question 19.
7 relatives of a man comprises 4 ladies and 3 gentlemen, his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways can they invite a dinner party of 3 ladies and 3 gentlemen so that there are 3 of man's relative and 3 of the wife's relatives?
Solution:

We need 3 ladies and 3 gentlemen for the party which consist of 3 Husbands relative and 3 wifes relative. This can be done as follows

$
\begin{aligned}
& { }^4 \mathrm{C}_0={ }^4 \mathrm{C}_4=1 ;{ }^3 \mathrm{C}_0={ }^3 \mathrm{C}_3=1 \\
& { }^4 \mathrm{C}_1={ }^4 \mathrm{C}_3=4 ;{ }^3 \mathrm{C}_1={ }^3 \mathrm{C}_2=3 \\
& { }^4 \mathrm{C}_2=\frac{4 \times 3}{2 \times 1}=6
\end{aligned}
$
(4) (1) (1) (4) $+(6)(3)(3)(6)+(4)(3)(3)(4)+(1)(1)(1)(1)=16+324+144+1=485$ ways
Question 20.
A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box, if at least one black ball is to be included in the draw?
Solution:
The box contains 2 white, 3 black and 4 red balls
We have to draw 3 balls in which there should be alteast 1 black ball
The possible draws are as follows
Black balls $=3$

Red and White $=2+4=6$
The possible ways are $\left(\begin{array}{l}3 \\ 1\end{array}\right)\left(\begin{array}{l}6 \\ 2\end{array}\right)+\left(\begin{array}{l}3 \\ 2\end{array}\right)\left(\begin{array}{l}6 \\ 1\end{array}\right)+\left(\begin{array}{l}3 \\ 3\end{array}\right)\left(\begin{array}{l}6 \\ 0\end{array}\right)$
$
\begin{aligned}
& { }^3 \mathrm{C}_1={ }^3 \mathrm{C}_2=3,{ }^3 \mathrm{C}_3=1 \\
& { }^6 \mathrm{C}_0=1 ;{ }^6 \mathrm{C}_1=6 ;{ }^6 \mathrm{C}_2=\frac{6 \times 5}{2 \times 1}=15
\end{aligned}
$
The possible ways are (3) $(15)+(3)(6)+(1)(1)=45+18+1=64$
Question 21.
Find the number of strings of 4 letters that can be formed with the letters of the word EXAMINATION.
Solution:
EXAMINATION
(i.e.) A, I, N are repeated twice. So the number of distinct letters $=8$
From the 8 letters we have to select and arrange 4 letters to form a 4 letter word which can
be done in ${ }^8 \mathrm{P}_4=8 \times 7 \times 6 \times 5=1680$
From the letters A, A, I, I, N, N when any 2 letters are taken as AA, II or AA, NN or II, NN
The number of 4 letter words $={ }^3 \mathrm{C}_2 \times \frac{4 !}{2 ! 2 !}$
(From II, AA, NN are select 2 sets)
(and we arrange the 4 letters) $=\frac{3 \times 24}{2 \times 2}=18$
From AA, II, NN we select one of them and from the remaining we select and arrange 3 which can be done in ways

$=3 \times \frac{7 \times 6}{2 \times 1} \times \frac{4 \times 3 \times 2}{2}=756$

$\therefore$ Total number of 4 letter word $=1680+18+756=2454$
Total number of 4 letter word $=1680+18+756=2454$
Question 22.
How many triangles can be formed by joining 15 points on the plane, in which no line joining any three points?
Solution:
No. of non-collinear points $=15$
To draw a Triangle we need 3 points
$\therefore$ Selecting 3 from 15 points can be done in ${ }^{15} \mathrm{C}_3$ ways.
$\therefore$ No. of Triangle formed $={ }^{15} \mathrm{C}_3=\frac{15 \times 14 \times 13}{3 \times 2 \times 1}=455$
Question 23.
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
Solution:
7 points lie on one line and the other 8 points parallel on another paraller line.
A triangle is obtained by taking one point from one line and second points from the other parallel line which can be done as follows.
$
\begin{aligned}
& { }^7 \mathrm{C}_1 \times{ }^8 \mathrm{C}_2 \text { or }{ }^7 \mathrm{C}_2 \times{ }^8 \mathrm{C}_1 \\
& { }^7 \mathrm{C}_1=7 ;{ }^7 \mathrm{C}_2=\frac{7 \times 6}{2 \times 1}=21 \\
& { }^8 \mathrm{C}_1=8 ;{ }^8 \mathrm{C}_2=\frac{8 \times 7}{2 \times 1}=28
\end{aligned}
$

$\therefore$ Number of triangles $=(7)(28)+(21)(8)=196+168=364$
Question 24 .
There are 11 points in a plane. No three of these lies in the same straight line except 4 points, which are collinear. Find,
(i) The number of straight lines that can be obtained from the pairs of these points?
Solution:
4 points are collinear

Total number of points 11 .
To get a line we need 2 points
$
\therefore \text { Number of lines }={ }^{11} \mathrm{C}_2=\frac{11 \times 10}{2 \times 1}=55
$
But in that 4 points are collinear
$\therefore$ We have to subtract ${ }^4 C_2=\frac{4 \times 3}{2 \times 1}=6$
From (1) Joining the 4 points we get 1 line
$
\therefore \text { Number of lines }={ }^{11} \mathrm{C}_2-{ }^4 \mathrm{C}_2+1=55-6+1=50
$
(ii) The number of triangles that can be formed for which the points as their vertices?
A triangle is obtained by joining 3 points.
So selecting 3 from 11 points can be
done in ${ }^{11} C_3=\frac{11 \times 10 \times 9}{3 \times 2 \times 1}=165$
But of the 11 points 4 points are collinear. So we have to subtract ${ }^4 C_3={ }^4 C_1=4$
$\therefore$ Number of triangles $=165-4=161$
Question 25.
A polygon has 90 diagonals. Find the number of its sides?
Solution:
A polygon with $n$ sides have ${ }^n C_2-n$ diagonals
Here ${ }^n \mathrm{C}_2-n=90$
$
\begin{aligned}
& \Rightarrow \quad \frac{n(n-1)}{2-1}-n=90 \\
& n^2-n-2 n=90 \times 2 \\
& n^2-3 n-180=0
\end{aligned}
$

$\begin{aligned}
& \quad \begin{array}{l}
(n-15)(n+12)=0 \\
n=15 \text { or }-12
\end{array} \\
& \text { but } n \neq-12 \\
& \therefore \mathrm{n}=15
\end{aligned}$