Exercise 3-Additional Questions - Chapter 4 Combinatorics and Mathematical Induction 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Questions
Question 1.
If $\frac{n !}{3 !(n-4) !}$ and $\frac{n !}{5 !(n-5) !}$ are in the ratio $5: 3$ find the value of $n$.
Solution:
$
\begin{aligned}
& \text { We have } \frac{n !}{3 !(n-4) !}: \frac{n !}{5 !(n-5) !}=5: 3 \\
& \Rightarrow \frac{n !}{3 !(n-4) !} \times \frac{5 !(n-5) !}{n !}=\frac{5}{3} \quad \Rightarrow \quad \frac{5 \times 4 \times 3 !(n-5) !}{3 !(n-4)(n-5) !}=\frac{5}{3} \Rightarrow \frac{20}{n-4}=\frac{5}{3} \\
& \Rightarrow n-4=20 \times \frac{3}{5} \quad \Rightarrow \quad n-4=12 \quad \Rightarrow n=16 \\
&
\end{aligned}
$

Question 2.
How many 3-digit even numbers can be made using the digits $1,2,3,4,6,7$ if no digit is repeated?
Solution:
Here total number of digits $=6$
The unit place can be filled with any one of the digits $2,4,6$.
So number of permutation $={ }^3 P_1=\frac{3 !}{2 !}=3$
Now the tens and hundreds place can be filled by remaining 5 digits.
So number of permutations $={ }^5 \mathrm{P}_3=\frac{5 !}{3 !}=\frac{5 \times 4 \times 3 !}{3 !}=20$
Hence total number of permutations $=3 \times 20=60$
Question 3.
Find $\mathrm{n}$ if ${ }^{\mathrm{n}-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 9$
Solution:
Here ${ }^{n-1} \mathrm{P}_3:{ }^n \mathrm{P}_4=1: 9$
$
\begin{aligned}
& \therefore \frac{(n-1) !}{(n-4) !}: \frac{n !}{(n-4) !}=1: 9 \\
& \Rightarrow \quad \frac{(n-1) !}{(n-4) !} \times \frac{(n-4) !}{n !}=\frac{1}{9} \quad \Rightarrow \frac{(n-1) !}{n !}=\frac{1}{9} \Rightarrow \frac{(n-1) !}{n(n-1) !}=\frac{1}{9} \\
& \Rightarrow \quad \frac{1}{n}=\frac{1}{9} \Rightarrow n=9
\end{aligned}
$

Question 4.
How many words can be formed by using the letters of the word ORIENTAL so that A ar E always occupy the odd places?
Solution:
[Hint: There are 4 odd places in the word]
$
\therefore \text { Number of permutations }={ }^4 P_2=\frac{4 !}{2 !}=12
$
Now the remaining 6 places filled with remaining 6 letters
$\therefore$ Number of permutations $={ }^6 \mathrm{P}_6=\frac{6 !}{0 !}=720$
Hence the total number of permutations $=12 \times 720=8640$
Question 5.
Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the points.
Solution:
Total number of points $=18$
Out of 18 numbers, 5 are collinear and we get a straight line by joining any two points.
$\therefore$ Total number of straight line formed by joining 2 points out of 18 points $={ }^{18} \mathrm{C}_2$
Number of straight lines formed by joining 2 points out of 5 points $={ }^5 \mathrm{C}_2$
But 5 points are collinear and we get only one line when they are joined pairwise.
So, the required number of straight lines are
$
={ }^{18} \mathrm{C}_2-{ }^5 \mathrm{C}_2+1=\frac{18 \cdot 17}{2 \cdot 1}-\frac{5 \cdot 4}{2 \cdot 1}+1=153-10+1=144
$
Hence, the total number of straight lines $=144$

Question 6.
We wish to select 6 person from 8 but, if the person $A$ is choosen, then B must be choosen. In how many ways can selections be made?
Solution:
Total number of persons $=8$
Number of persons to be selected $=6$
Condition is that if $\mathrm{A}$ is choosen, $\mathrm{B}$ must be choosen
Case I: When A is choosen, B must be choosen
Number of ways $={ }^6 \mathrm{C}_4$
[ $\therefore \mathrm{A}$ and $\mathrm{B}$ are set to be choosen]

Case II: When A is not choosen, then B may be choosen
$
\begin{aligned}
\therefore & \text { Number of ways }={ }^7 \mathrm{C}_6 \\
& ={ }^6 \mathrm{C}_2+{ }^7 \mathrm{C}_1 \\
& \left.=\frac{6.5}{2.1}+7=15+7=22 \text { where are two case }{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{n-r}\right]
\end{aligned}
$
Hence, the required number of ways $=22$
Question 7.
How many 3-digit even numbers can be made using the digits $1,2,3,4,6,7$, if no digit is repeated?
Solution:
For 3-digit even numbers, the unit's place can be occupied by one of the 3 digits 2,4 or 6 . The remaining 5 digits can be arranged in the remaining 2 places in ${ }^5 \mathrm{P}_2$ ways.

Question 8.
Find the number of 4-digit numbers that can be formed using the digits 1, 2,3,4, 5 if no digit is repeated. How many of these will be even?
Solution:

For 4 digit numbers, we have to arrange the given 5 digits in 4 vacant places. This can be done in ${ }^5 \mathrm{P}_4=5 \times 4 \times 3 \times 2=120$ ways.

(i) ${ }^5 \mathrm{P}_{\mathrm{r}}=2{ }^6 \mathrm{P}_{\mathrm{r}-1} 4$
Solution:
$
\begin{aligned}
& \Rightarrow \quad \frac{5 !}{(5-r) !}=2 \times \frac{6 !}{\{6-(r-1)\} !} \\
& \Rightarrow \quad \frac{5 !}{(5-r) !}=2 \times \frac{6.5 !}{(7-r) !} \\
& \Rightarrow \quad \frac{5 !}{(5-r) !}=\frac{12}{(7-r)(6-r)(5-r) !} \\
& 1=\frac{12}{(7-r)(6-r)} \\
& \Rightarrow r^2-13 r+42=12 \quad \text { or } r^2-13 r+30=0 \\
& \text { or }(r-3)(r-10)=0 \\
& \therefore r=3,10 \\
&
\end{aligned}
$
Now, ${ }^5 \mathrm{P}_r$ and ${ }^6 \mathrm{P}_{r-1}$ are meaningless when $r=10$
$\left[\therefore n_r\right.$ is defined only if $\left.r \leq n\right]$
$\therefore$ Rejecting $r=10$, we have $r=3$

$
\begin{aligned}
& \text { (ii) }{ }^5 \mathrm{P}_{\mathrm{r}}={ }^6 \mathrm{P}_{\mathrm{r}-1} \\
& \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{5 !}{(5-r) !}=\frac{6}{(7-r)(6-r)(5-r) !} \\
& \Rightarrow \quad 1=\frac{6}{(7-r)(6-r)} \quad \Rightarrow(7-r)(6-r)=6 \\
& \Rightarrow 42-13 r+r^2=6 \Rightarrow r^2-13 r+36=0 \\
& \Rightarrow(\mathrm{r}-4)(\mathrm{r}-9)=0 \Rightarrow \mathrm{r}=4,9 \\
&
\end{aligned}
$
Now, we know that ${ }^n P_r$ is meaningful only when $r \leq n$.
$\therefore{ }^5 \mathrm{P}_{\mathrm{r}}$ and ${ }^6 \mathrm{P}_{\mathrm{r}-1}$ are meaningless when $\mathrm{r} \leq 9$.
$\therefore$ Rejecting $r=9$, we have $r=4$

Question 10.
How many words, with or without meaning, can be made from the letters of the word MONDAY, assuming that no letter is repeated, if

(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used but first letter is a vowel?
Solution:
The word MONDAY has 6 distinct letters.
(i) 4 letters out of 6 can be arranged in ${ }^6 \mathrm{P}_4$ ways.
$\therefore$ 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.
$\therefore$ The required number of words $={ }^6 \mathrm{P}_6=6$ !
$=1 \times 2 \times 3 \times 4 \times 5 \times 6=720$.
(iii) The first place can be filled by anyone $O f$ 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 ${ }^5 \mathrm{P}_5=5$ ! ways.
$\therefore$ By the multiplication rule, the required number of words $=2 \times 5 !=2 \times 1 \times 2 \times 3 \times 4 \times 5$
$=240$