Exercise 4.3-Additional Questions - Chapter 4 - Combinatorics and Mathematical Induction - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Questions Solved
Question 1.
A group consists of 4 girls and 7 boys. In bow many ways can a team of 5 members be selected, if the team has
(i) no girls
(ii) atleast one boy and one girl.
(iii) at least three girls
Solution:
We have 4 girls and 7 boys and a team of 5 members is to be selected.
(i) If no girl in selected, then all the 5 members are to be selected out of 7 boys
(i.e) ${ }^7 C_5=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6.5 !}{5 ! \times 2}=21$ ways
(ii) When at least one boy and one girl are to be selected, then
(i.e) ${ }^7 C_5=\frac{7 !}{5 ! 2 !}=\frac{7 \times 6.5 !}{5 ! \times 2}=21$ ways
(ii) When at least one boy and one girl are to be selected, then
$
\begin{aligned}
\text { Number of ways } & ={ }^4 \mathrm{C}_1 \times{ }^7 \mathrm{C}_4+{ }^4 \mathrm{C}_2 \times{ }^7 \mathrm{C}_3+{ }^4 \mathrm{C}_3 \times{ }^7 \mathrm{C}_2+{ }^4 \mathrm{C}_4 \times{ }^7 \mathrm{C}_1 \\
& =4 \times \frac{7 \times 6 \times 5 \times 4}{4 \times 3 \times 2 \times 1}+\frac{4 \times 3}{2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}+4 \times \frac{7 \times 6}{2 \times 1}+1 \times 7 \\
& =(4 \times 35)+(6 \times 35)+(4 \times 21)+7=140+102+84+7=441 \text { way }
\end{aligned}
$
(iii) When atleast 3 girls are included, then
$
\begin{aligned}
& \text { Number of ways }={ }^4 \mathrm{C}_3 \times{ }^7 \mathrm{C}_2+{ }^4 \mathrm{C}_4 \times{ }^7 \mathrm{C}_1 \\
& =4 \times \frac{7 \times 6}{2 \times 1}+1 \times 7=84+7=91 \text { ways }
\end{aligned}
$
Hence the required number of ways are (i) 21 ways (ii) 441 ways (iii) 91 ways

Question 2.
A committee of 6 is to be choosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done it two particular women refuse to serve on the same committee?
Solution:
We have 10 men and 7 women out of which a committee of 6 is to be formed which contain atleast 3 men and 2 women
$
\begin{aligned}
& \text { Therefore, Number of ways }={ }^{10} \mathrm{C}_3 \times{ }^7 \mathrm{C}_3+{ }^{10} \mathrm{C}_4 \times{ }^7 \mathrm{C}_2 \\
& =\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times \frac{7 \times 6 \times 5}{3 \times 2 \times 1}+\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times \frac{7 \times 6}{2 \times 1} \\
& =(120 \times 35)+(210 \times 21)=4200+4410=8610 \\
&
\end{aligned}
$
If 2 particular women to be always present, then the number of ways
$
\begin{aligned}
& ={ }^{10} \mathrm{C}_4 \times{ }^5 \mathrm{C}_0+{ }^{10} \mathrm{C}_3 \times{ }^5 \mathrm{C}_1 \\
& =\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \times 1+\frac{10 \times 9 \times 8}{3 \times 2 \times 1} \times 5=210+120 \times 5=210+600=810
\end{aligned}
$
$\therefore$ Total number of committee $=8610-810=7800$
Hence, the value of the filler is 7800
Question 3.
Using the digits $1,2,3,4,5,6,7$ a number of 4 different digits is formed. Find

Solution:
(a) Total of 4 digit number formed with $1,2,3,4,5,6,7$
$
={ }^7 \mathrm{P}_4=\frac{7 !}{(7-4) !}=\frac{7 \times 6 \times 5 \times 4 \times 3 !}{3 !}=840
$
(b) When a number is divisible by $2=4 \times 5 \times 6 \times 3=360$
(c) Total numbers which are divisible by $25=40$
(d) Total numbers which are divisible by 4 (last two digits is divisble by 4 ) $=200$ Hence, the required matching is (a) $\leftrightarrow$ (z), (b) $\leftrightarrow$ (iii), (c) $\leftrightarrow$ (iv), (d) $\leftrightarrow$ (ii)
Question 4

 If ${ }^{22} \mathrm{P}_{\mathrm{r}+1}:{ }^{20} \mathrm{P}_{\mathrm{r}+2}=11: 52$, find $\mathrm{r}$
Solution:
Here ${ }^{22} \mathrm{P}_{r+1}:{ }^{20} \mathrm{P}_{r+2}=11: 52$
$
\Rightarrow \frac{22 !}{(21-r) !} \times \frac{(18-r) !}{20 !}=\frac{11}{52}
$
$
\begin{aligned}
& \Rightarrow \frac{22 \times 21 \times 20 !}{(21-r)(20-r)(19-r)(18-r) !} \times \frac{(18-r) !}{20 !}=\frac{11}{52} \\
& \Rightarrow \frac{22 \times 21}{(21-r)(20-r)(19-r)}=\frac{11}{52} \\
& \Rightarrow(21-r)(20-r)(19-r)=2 \times 21 \times 52 \\
& \Rightarrow(21-r)(20-r)(19-r)=14 \times 13 \times 12 \\
& \Rightarrow(21-r)(20-r)(19-r)=(21-7)(20-7)(19-7) \\
& \Rightarrow r=7
\end{aligned}
$
Question 5 .
If ${ }^n \mathrm{P}_r={ }^n \mathrm{P}_{r+1}$ and ${ }^n \mathrm{C}_r={ }^n \mathrm{C}_{r-1}$, find the values of $n$ and $r$.

Solution:
Here ${ }^n \mathrm{P}_r={ }^n \mathrm{P}_{r+1}$
$
\begin{aligned}
& \Rightarrow \frac{n !}{(n-r) !}=\frac{n !}{(n-r-1) !} \\
& \Rightarrow \frac{1}{(n-r)(n-r-1)}=\frac{1}{(n-r-1) !} \\
& \Rightarrow \frac{1}{n-r}=1 \Rightarrow n-r=1 \\
& \text { Also }{ }^n \mathrm{C}_r={ }^n \mathrm{C}_{r-1} \\
& \Rightarrow \frac{n !}{(n-r) ! r !}=\frac{n !}{(n-r+1) !(r-1) !} \\
& \Rightarrow \frac{1}{(n-r) ! r(r-1) !}=\frac{1}{(n-r+1)(n-r) !(r-1) !} \\
& \Rightarrow \frac{1}{r}=\frac{1}{n-r+1} \\
& \Rightarrow n-r+1=r \Rightarrow n-2 r=-1 \\
&
\end{aligned}
$
From $(i)$ and $(i i)$, we have $n=3$ and $r=2$
Question 6.
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls?
(ii) at least 3 girls?
(iii) almost 3 girls?
Solution:
(i) There are 9 boys and 4 girls. We have to select exactly 3 girls out of 4 girls and 4 boys $\begin{aligned} & \text { out of } 9 \text { boys. } \\ & \therefore \text { Number of ways of selection }={ }^9 \mathrm{C}_4 \times{ }^4 \mathrm{C}_3=\frac{9 !}{4 ! 5 !} \times \frac{4 !}{3 ! 1 !} \\ &=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 ! 5 !} \times \frac{4 !}{3 \times 2 \times 1}=504\end{aligned}$
(ii) We have to select at least 3 girls. So the committee consists of 3 girls and 4 boys or 4 girls and 3 boys.

$
\begin{aligned}
\therefore & \text { Number of ways of selection }={ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4+{ }^4 \mathrm{C}_4 \times{ }^9 \mathrm{C}_3 \\
& =\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !}+\frac{4 !}{4 ! 0 !} \times \frac{9 !}{3 ! 6 !} \\
& =\frac{4 !}{3 \times 2 \times 1} \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 ! 5 !}+1 \times \frac{9 \times 8 \times 7 \times 6 !}{3 \times 2 \times 1 \times 6 !} \\
& =504+84=588
\end{aligned}
$
(iii) We have to select at most 3 girls. So the committee consists of no girl and 7 boys or 1 girl and 6 boys or 2 girls and 5 boys or 3 girls and 4 boys.
$
\begin{aligned}
\therefore \text { Number of ways of selection }= & { }^4 \mathrm{C}_0 \times{ }^9 \mathrm{C}_7+{ }^4 \mathrm{C}_1 \times{ }^9 \mathrm{C}_6+{ }^4 \mathrm{C}_2 \times{ }^9 \mathrm{C}_5+{ }^4 \mathrm{C}_3 \times{ }^9 \mathrm{C}_4 \\
= & 1 \times \frac{9 !}{7 ! 2 !}+\frac{4 !}{1 ! 3 !}+\frac{9 !}{6 ! 3 !}+\frac{4 !}{2 ! 2 !} \times \frac{9 !}{5 ! 4 !}+\frac{4 !}{3 ! 1 !} \times \frac{9 !}{4 ! 5 !} \\
= & 1 \times \frac{9 \times 8 \times 7 !}{7 ! \times 2 \times 1}+\frac{4 \times 3 !}{1 \times 3 !} \times \frac{9 \times 8 \times 7 \times 6 !}{6 ! \times 3 \times 2 \times 1} \\
& +\frac{4 \times 3 \times 2 !}{2 \times 1 \times 2 !} \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 ! \times 4 \times 3 \times 2 \times 1} \\
& +\frac{4 \times 3 !}{3 ! \times 1} \times \frac{9 \times 8 \times 7 \times 6 \times 5 !}{4 \times 3 \times 2 \times 1 \times 5 !} \\
= & 36+336+756+504=1632
\end{aligned}
$
Question 7.
Determine $\mathrm{n}$ if
(i) ${ }^{2 \mathrm{n}} \mathrm{C}_3:{ }^{\mathrm{n}} \mathrm{C}_2=12: 1$
(ii) ${ }^{2 \mathrm{n}} \mathrm{C}_3:{ }^{\mathrm{n}} \mathrm{C}_3=11: 1$
Solution:

(i) Here ${ }^{2 n} \mathrm{C}_3:{ }^n \mathrm{C}_2=12: 1$
$
\begin{aligned}
& \Rightarrow \quad \frac{(2 n) !}{3 !(2 n-3) !} \times \frac{2 !(n-2) !}{n !}=\frac{12}{1} \\
& \Rightarrow \quad \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{3 \times 2 !(2 n-3) !} \times \frac{2 !(n-2) !}{n(n-1)(n-2) !}=\frac{12}{1} \\
& \Rightarrow \quad \frac{(2 n)(2 n-1)(2 n-2)}{3} \times \frac{1}{n(n-1)}=\frac{12}{1} \\
& \Rightarrow \quad \frac{4(2 n-1)}{3}=\frac{12}{1} \\
& \Rightarrow 8 n-4=36 \Rightarrow n=5
\end{aligned}
$

(ii) Here ${ }^{2 n} \mathrm{C}_3:{ }^n \mathrm{C}_3=11: 1$
$
\begin{aligned}
& \Rightarrow \quad \frac{(2 n) !}{3 !(2 n 3) !} \times \frac{2 !(n 3) !}{n !}=\frac{11}{1} \\
& \Rightarrow \quad \frac{(2 n)(2 n-1)(2 n-2)(2 n-3) !}{3 !(2 n-3) !} \times \frac{3 !(n-3) !}{n(n-1)(n-2)(n-3) !}=\frac{11}{1} \\
& \Rightarrow \quad \frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)}=\frac{11}{1} \\
& \Rightarrow \quad \frac{4(2 n-1)}{n-2}=\frac{11}{1} \\
& \Rightarrow \quad 8 n-4=11 n-22 \\
& \Rightarrow \quad 3 n=18 \Rightarrow n=6
\end{aligned}
$
Question 8.
Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
Solution:
There are 6 red balls, 5 white balls and 5 blue balls.
We have to select 3 balls of each colour.
$
\begin{aligned}
\therefore \text { Number of ways of selection } & ={ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3 \\
& =\frac{6 !}{3 ! 3 !} \times \frac{5 !}{3 ! 2 !} \times \frac{5 !}{3 ! 2 !} \\
& =20 \times 10 \times 10=2000
\end{aligned}
$

Question 9.
If $\frac{1}{7 !}+\frac{1}{9 !}=\frac{x}{10 !}$, find $x$.
Solution:
Here $\frac{1}{7 !}+\frac{1}{9 !}=\frac{x}{10 !}$
$
\begin{aligned}
\Rightarrow & \frac{1}{7 !}+\frac{1}{9 \times 8 \times 7 !} & =\frac{x}{10 \times 9 \times 8 \times 7 !} \\
\Rightarrow & \frac{1}{7 !}\left[1+\frac{1}{72}\right] & =\frac{1}{7 !}\left[\frac{x}{10 \times 9 \times 8}\right] \\
\Rightarrow & \frac{73}{72} & =\frac{x}{10 \times 9 \times 8} \\
\Rightarrow & x & =\frac{73}{72} \times 10 \times 9 \times 8=730
\end{aligned}
$

Question 10 .
If ${ }^{\mathrm{n}} \mathrm{C}_4,{ }^{\mathrm{n}} \mathrm{C}_5$ and ${ }^{\mathrm{n}} \mathrm{C}_4$ are in A.P. then find $\mathrm{n}$.
[Hint: $\left.{ }^{2 \mathrm{n}} \mathrm{C}_5={ }^{\mathrm{n}} \mathrm{C}_6+{ }^{\mathrm{n}} \mathrm{C}_4\right]$
Solution:
$
\begin{aligned}
& 2 \frac{n !}{(n-5) ! 5 !}=\frac{n !}{(n-6) ! 6 !}+\frac{n !}{(n-4) ! 4 !} \\
& \Rightarrow \quad \frac{2}{(n-5)(n-6) ! .5 \cdot 4 !}=\frac{1}{(n-6) ! \cdot 6 \cdot 5 \cdot 4 !}+\frac{1}{(n-4)(n-5)(n-6) ! 4 !} \\
& \Rightarrow \frac{2}{5(n-5)}=\frac{1}{30}+\frac{1}{(n-4)(n-5)} \Rightarrow \frac{2}{5(n-5)}-\frac{1}{(n-4)(n-5)}=\frac{1}{30} \\
& \Rightarrow \quad \frac{1}{n-5}\left[\frac{2}{5}-\frac{1}{n-4}\right]=\frac{1}{30} \Rightarrow \frac{1}{n-5}\left[\frac{2 n-8-5}{5(n-4)}\right]=\frac{1}{30} \\
& \Rightarrow \quad \frac{2 n-13}{5(n-4)(n-5)}=\frac{1}{30} \Rightarrow \frac{2 n-13}{n_2^2-9 n+20}=\frac{1}{6} \\
& \Rightarrow \quad 12 n-78=n^2-9 n+20 \Rightarrow n^2-21 n+98=0 \\
& \Rightarrow \quad(n-14)(n-7)=0 \Rightarrow n=14 \text { or } 7 \\
&
\end{aligned}
$