Exercise 4.5 - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Ex 4.5
Question 1.
The sum of the digits at the 10th place of all numbers formed with the help of 2,4 , 5,7 taken all at a time is.$\ldots \ldots$.
(a) 432
(b) 108
(c) 36
(d) 18
Solution:
(b) 108
Hint. Number of digits given $=2,4,5,7$
Number of 4 digit numbers formed $=4 !=24$
So each digit occur $\frac{24}{4}=6$ times
Sum of the digits $=2+4+5+7=18$
So sum of the digits in each place $=18 \times 6=108$
Question 2.
In an examination there are three multiple choice questions and each question has 5 choices. Number of ways in which a student can fail to get all answer correct is
(a) 125
(b) 124
(c) 64
(d) 63

Solution:
(b) 124
Hint. Each question has 5 options in which 1 is correct
So the number of ways of getting correct answer for all the three questions is $5^3=$ 125

So the number of ways in which a student can fail to get all answer correct is $<125$ (i.e.) $125-1=124$
Question 3.
The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is
(a) $30^4 \times 29^2$
(b) $30^3 \times 29^3$
(c) $30^2 \times 29^4$
(d) $30 \times 29^5$
Solution:
(a) $30^4 \times 29^2$
Hint.
I and II in maths can be given can be given in $30 \times 29$ ways.
I and II in physics can be given in $30 \times 29$ ways.
I and chemistry can be given in 30 ways.
I in English can be given in 30 ways.
So total number of ways $=30 \times 29 \times 30 \times 29 \times 30 \times 30=30^4 \times 29^2$

Question 4.
The number of 5 digit numbers all digits of which are odd is
(a) 25
(b) $5^5$
(c) $5^6$
(d) 625
Solution:
(b) $5^5$
Hint. The odd number are $1,3,5,7,9$
Number of odd numbers $=5$
We need a five digit number So the number of five digit number $=5^5$
Question 5.
In 3 fingers, the number of ways four rings can be worn is ways.

(a) $4^3-1$
(b) $3^4$
(c) 68
(d) 64
Solution:
(b) $3^4$
Hint. Each letter can be ported in 3 ways
$\therefore 4$ letter is $3^4$ ways
Question 6.
If $^{(n+5)} \mathrm{P}_{(n+1)}=\left(\frac{11(n-1)}{2}\right)^{(n+3)} \mathrm{P}_n$, then the value of $n$ are
(a) 7 and 11
(b) 6 and 7
(c) 2 and 11
(d) 2 and 6
Solution:
(b) 6 and 7
Question 7.
The product of $r$ consecutive positive integers is divisible by
(a) $\mathrm{r}$ !
(b) $(\mathrm{r}-1)$ !
(c) $(r+1)$ !

(d) $r$ !
Solution:
(a) $r$ !
Hint.
1(2) (3) .....(r) $=r$ ! which is $\div$ by $r$ !
Question 8.
The number of 5 digit telephone numbers which have none of their digits repeated is
(a) 90000
(b) 10000
(c) 30240
(d) 69760
Solution:
(d) 69760
Hint.
The number of 5 digit telephone numbers which have none of their digits repeated

is ${ }^{10} \mathrm{P}_5=30240$
Thus the required number of telephone number is $10^5-30240=69760$
Question 9.
If $\mathrm{a}^2-{ }^{\mathrm{a}} \mathrm{C}_2=\mathrm{a}^2-{ }^{\mathrm{a}} \mathrm{C}_4$ then the value of ' $\mathrm{a}$ ' is ....
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(b) 3
Hint.
$
\begin{aligned}
& a_2-a=2+4=6 \\
& a_2-a-6=0 \\
& (a-3)(a+2)=0 \Rightarrow a=3
\end{aligned}
$
Question 10.
There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is .......
(a) 45
(b) 40
(c) 39
(d) 38

Solution:
(b) 40
Hint.
$
\begin{aligned}
{ }^{10} C_2-{ }^4 C_2+1 & =\frac{10 \times 9}{2 \times 1}-\frac{4 \times 3}{2 \times 1}+1 \\
& =45-6+1=40
\end{aligned}
$
Question 11.
The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is

(a) $2 \times{ }^{11} \mathrm{C}_7+{ }^{10} \mathrm{C}_8$
(b) ${ }^{11} \mathrm{C}_7+{ }^{10} \mathrm{C}_8$
(c) ${ }^{12} \mathrm{C}_8-{ }^{10} \mathrm{C}_6$
(d) ${ }^{10} \mathrm{C}_6+2$ !
Solution:
(c) ${ }^{12} \mathrm{C}_8-{ }^{10} \mathrm{C}_6$
Hint.
Number of way of selecting 8 people from 12 in ${ }^{12} \mathrm{C}_8$
$\therefore$ out of remaining people 8 can attend in ${ }^{10} \mathrm{C}_8$
The number of ways in which two of them do not attend together $={ }^{12} \mathrm{C}_8-{ }^{10} \mathrm{C}_6$
Question 12.
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines .......
(a) 6
(b) 9
(c) 12
(d) 18
Solution:
(d) 18
Hint.
Number of parallelograms $={ }^4 \mathrm{C}_2 \times{ }^3 \mathrm{C}_2$
$
=6 \times 3=18
$

Question 13.
Everybody in a room shakes hands with everybody else. The total number of shake hands is 66 . The number of persons in the room is
(a) 11
(b) 12
(c) 10
(d) 6
Solution:
(b) 12
Hint.
Number of shake hands $=\frac{n(n-1)}{2 \times 1}=66 ; n(n-1)=132=12 \times 11 \Rightarrow n=12$

Question 14.
Number of sides of a polygon having 44 diagonals is
(a) 4
(b) 4 !
(c) 11
(d) 22
Solution:
(c) 11
Hint:
$
\begin{aligned}
& { }^n \mathrm{C}_2-n=44 \\
& \frac{n(n-1)}{2 \times 1}-n=44 \\
& \frac{n^2-n-2 n}{2}=44 ; n^2-3 n-88=0 \\
& (n-11)(n+8)=0 \quad \Rightarrow \quad n=11 \text { or }-8 \\
& \therefore n=11 \quad n \neq-8 \\
&
\end{aligned}
$
Question 15.
If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are
(a) 45
(b) 40
(c) 10 !

(d) $2^{10}$
Solution:
(a) 45
Hint:
$
{ }^{10} C_2=\frac{10 \times 9}{2 \times 1}=45
$
Question 16.
In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is
(a) 110
(b) ${ }^{10} \mathrm{C}_3$
(c) 120
(d) 116
Solution:

(d) 116
Hint:
Number of triangles $={ }^{10} \mathrm{C}_3-{ }^4 \mathrm{C}_3=\frac{10 \times 9 \times 8}{3 \times 2 \times 1}-4=120-4=116$
Question 17.
In ${ }^{2 \mathrm{n}} \mathrm{C}_3:{ }^n \mathrm{C}_3=11: 1$ then $\mathrm{n}$ is
(a) 5
(b) 6
(c) 11
(d) 1
Solution:
(b) 6

Hint.
$
\begin{aligned}
& \frac{{ }^{2 n} \mathrm{C}_3}{{ }^n \mathrm{C}_3}=\frac{11}{1} \Rightarrow \frac{2 n !}{3 !(2 n-3) !} / \frac{n !}{3 !(n-3) !}=11 \\
& \frac{2 n !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n !}=11 \\
& \frac{2 n(2 n-1)(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) !} \times \frac{(n-3) !}{n(n-1)(n-2)(n-3) !}=11 \\
& \frac{2 n(2 n-1)(2 n-2)}{n(n-1)(n-2)}=11 \\
& \frac{2 n(2 n-1) 2(n-1)}{n(n-1)(n-2)}=11 \\
& 4(2 n-1)=11(n-2) \Rightarrow 8 n-4=11 n-22 \\
& 18=3 n \Rightarrow n=6
\end{aligned}
$

Question 18.
${ }^{(n-1)} C_r+{ }^{(n-1)} C_{(r-1)}$ is
(a) ${ }^{(\mathrm{n}+1)} \mathrm{C}_{\mathrm{r}}$
(b) ${ }^{(n-1)} \mathrm{C}_{\mathrm{r}}$
(c) ${ }^n \mathrm{C}_{\mathrm{r}}$
(d) ${ }^n C_{r-1}$
Solution:
(c) ${ }^n \mathrm{C}_{\mathrm{r}}$
$
\left(\begin{array}{c}
n-1 \\
r
\end{array}\right)+\left(\begin{array}{c}
n-1 \\
r-1
\end{array}\right)=\left(\begin{array}{l}
n \\
r
\end{array}\right)
$
Question 19.
The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king is

(a) ${ }^{52} \mathrm{C}_5$
(b) ${ }^{48} \mathrm{C}_5$
(c) ${ }^{52} \mathrm{C}_5+{ }^{48} \mathrm{C}_5$
(d) ${ }^{52} \mathrm{C}_5-{ }^{48} \mathrm{C}_5$
Solution:
(d) ${ }^{52} \mathrm{C}_5-{ }^{48} \mathrm{C}_5$
Hint.
Selecting 5 from 52 cards $={ }^{52} \mathrm{C}_5$
selecting 5 from the (non-king cards 48 ) $={ }^{48} \mathrm{C}_5$
$\therefore$ Number of ways is ${ }^{52} \mathrm{C}_5-{ }^{48} \mathrm{C}_5$
Question 20.
The number of rectangles that a chessboard has ......
(a) 81
(b) 99
(c) 1296
(d) 6561
Solution:
(c) 1296
Hint. Number of horizontal times $=9$
Number of vertical times $=9$
Selecting 2 from 9 horizontal lines $={ }^9 \mathrm{C}_2$

Selecting 2 from 9 horizontal lines $={ }^9 \mathrm{C}_2$
Selecting 2 from 9 vertical lines $={ }^9 \mathrm{C}_2$
$
{ }^9 C_2 \times{ }^9 C_2=\frac{9 \times 8}{2 \times 1} \times \frac{9 \times 8}{2 \times 1}=36 \times 36=1296
$
Question 21.
The number of 10 digit number that can be written by using the digits 2 and 3 is
(a) ${ }^{10} \mathrm{C}_2+{ }^9 \mathrm{C}_2$
(b) $2^{10}$
(c) $2^{10}-2$
(d) 10 !
Solution:
(b) $2^{10}$

Hint.
Selecting the number from (2 and 3$)$
For till the first digit can be done in 2 ways
For till the second digit can be done in 2 ways ....
For till the tenth digit can be done in 2 ways
So, total number of ways in 10 digit number $=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$ $=2^{10}$
Question 22.
If $\mathrm{P}_{\mathrm{r}}$ stands for ${ }^{\mathrm{r}} \mathrm{P}_{\mathrm{r}}$ then the sum of the series $1+\mathrm{P}_1+2 \mathrm{P}_2+3 \mathrm{P}_3+\ldots+n \mathrm{P}_{\mathrm{n}}$ is
(a) $P_{n+1}$
(b) $\mathrm{P}_{\mathrm{n}+1}-1$
(c) $\mathrm{P}_{\mathrm{n}-1}+1$
(d) ${ }^{(n+1)} P_{(n-1)}$
Solution:
(b) $\mathrm{P}_{\mathrm{n}+1}-1$
Hint:
$1+1 !+2 !+3 !+\ldots+n$ !
Now $1+1(1 !)=2=(1+1)$ !
$1+1(1 !)+2(2 !)=1+1+4=6=3$ !
$1+1(1 !)+2(2 !)+3(3 !)=1+1+4+18=24=4$ !
$1+1(1 !)+2(2 !)+3(3 !) \ldots+n(n !)=(n+1) !-1$

$
={ }^{n+1} P_{n+1}-1=P_{n+1}-1
$
Question 23.
The product of first $\mathrm{n}$ odd natural numbers equals ......
(a) ${ }^{2 n} \mathrm{C}_n \times{ }^n \mathrm{P}_n$
(b) $\left(\frac{1}{2}\right)^n \times{ }^{2 n} \mathrm{C}_n \times{ }^n \mathrm{P}_n$
(c) $\left(\frac{1}{4}\right)^n \times{ }^{2 n} \mathrm{C}_n \times{ }^{2 n} \mathrm{P}_n$
(d) ${ }^n \mathrm{C}_n \times{ }^n \mathrm{P}_n$
Solution:
(b) $\left(\frac{1}{2}\right)^n \times{ }^{2 n} \mathrm{C}_n \times{ }^n \mathrm{P}_n$

Hint:
$
\begin{gathered}
1.3 .5 \ldots(2 n-1)=\frac{1.2 .3 .4 \ldots(2 n-1)(2 n)}{2.4 \ldots(2 n)} \\
=\frac{2 n !}{2^n n !}=\left(\frac{1}{2}\right)^n{ }^{2 n} \mathrm{C}_n \times{ }^n \mathrm{P}_n
\end{gathered}
$
Question 24.
If ${ }^{\mathrm{n}} \mathrm{C}_4,{ }^{\mathrm{n}} \mathrm{C}_5,{ }^{\mathrm{n}} \mathrm{C}_6$ are in AP the value of $\mathrm{n}$ can be
(a) 14
(b) 11
(c) 9
(d) 5
Solution:
(a) 14
Hint:
$
\begin{aligned}
& { }^n \mathrm{C}_4,{ }^n \mathrm{C}_5,{ }^n \mathrm{C}_6 \text { are in A.P } \\
& \Rightarrow \quad 2\left[{ }^n \mathrm{C}_5\right]={ }^n \mathrm{C}_4+{ }^n \mathrm{C}_6 \\
& \Rightarrow \quad \frac{2(n !)}{5 !(n-5) !}=\frac{n !}{4 !(n-4) !}+\frac{n !}{6 !(n-6) !}
\end{aligned}
$

$
\begin{aligned}
& \frac{n !}{6 !(n-5) !}[2(6)(n-4)=5 \times 6+(n-5)(n-4)] \\
& 12(n-4)=30+n^2-9 n+20 \\
& 30+\mathrm{n}^2-9 \mathrm{n}+20-12 \mathrm{n}+48=0 \\
& \mathrm{n}^2-21 \mathrm{n}+98=0 \\
& (\mathrm{n}-7)(\mathrm{n}-14)=0 \\
& \mathrm{n}=7(\text { or }) 14
\end{aligned}
$
Question 25 .
$1+3+5+7++17$ is equal to
(a) 101
(b) 81
(c) 71
(d) 61

Solution:
(b) 81
Hint:
$
1+3+5+\ldots+17=\left(\frac{17+1}{2}\right)^2=9^2=81
$