2 Marks Questions

Question 12 Marks

A committee of 6 members is to be formed out of 8 men and 5 women. In how many ways can this committee be formed while in each committee :
(i) there should be only 2 men,
(ii) there should be only 2 women,
(iii) there should be at least two women,
(iv) there should be at least two men.

Answer

(i) Out of 8 men, only 2 men can be selected by ${ }^8 C _2$ method and remaining 4 are to be selected from 5 women. Who can be selected by ${ }^5 C _4$. Hence, required number
$={ }^8 C_2 \times{ }^5 C_4=28 \times 5=140$
(ii) Out of 5 women, only 2 women can be selected from ${ }^5 C _2$ type and remaining 4 are to be selected from 8 men who can be selected from ${ }^8 C _4$ type.
Hence, required number
$={ }^5 C_2 \times{ }^8 C_4=10 \times 70=700$
(iii) To select atleast two women, selection can be made in the following manner :
2 women and 4 men, 3 women and 3 men, 4 women and 2 men, 5 women and 1 man.
Hence, their selection will be as follows :
${ }^5 C_2 \times{ }^8 C_4,{ }^5 C_3 \times{ }^8 C_3,{ }^5 C_4 \times{ }^8 C_2,{ }^5 C_5 \times{ }^8 C_1$
Hence the required number
$\begin{array}{l}={ }^5 C_2 \times{ }^8 C_4+{ }^5 C_3 \times{ }^8 C_3+{ }^5 C_4 \times{ }^8 C_2+{ }^5 C_5 \times{ }^8 C_1 \\=70+560+140+8=1408\end{array}$
(iv) By the above mentioned method, the total methods of making committees of 6 members with at least 2 men :
$\begin{array}{l}{ }^8 C_2 \times{ }^5 C_4+{ }^8 C_3 \times{ }^5 C_3+{ }^8 C_4 \times{ }^5 C_2+{ }^8 C_5 \times{ }^5 C_1+{ }^8 C_6 \\=140+560+700+280+28=1708\end{array}$

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Question 22 Marks

In how many ways can 11 players of cricket team be selected out of 14 players? Here often a particular player
(i) is always included,
(ii) never selected.

Answer

Ways to select 11 players out of 14 players
$\begin{array}{l}\quad{ }^{14} C_{11} \\=\frac{14!}{11!(14-11)!}=\frac{14!}{11!3!} \\=\frac{14 \times 13 \times 12}{11!3!} 11!=\frac{14 \times 13 \times 12}{3 \times 2 \times 1}=364\end{array}$
(i) When a particular player has to be included in the team always, then 10 players have to be selected from the remaining 13 players. Hence the required ways of selecting 10 out of 13 will be
$\begin{array}{l}={ }^{13} C_{10}=\frac{13!}{10!(13-10)!}=\frac{13!}{10!3!} \\=\frac{13 \times 12 \times 11}{3 \times 2 \times 1}=286\end{array}$
(ii) When a particular player is never to be selected then 11 players have to be selected from the remaining 13 players. Hence, the required methods in this situation would be
$\begin{array}{l}={ }^{13} C_{11}=\frac{13!}{11!(13-11)!} \\=\frac{13!}{11!2!}=\frac{13 \times 1211!}{11!2!}=78\end{array}$

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Question 32 Marks

How many words can be made from the letters of the word DELHI, if
(i) All letters should be taken
(ii) Only 3 letters should be taken
(iii) Every word starts with D
(iv) Every word starts with D and ends with I
(v) Both the vowels come together
(vi) Every word starts with both vowels?

Answer

All letters in DELHI are different.
(i) Number of permutations made using all 5 letters will be
$\begin{array}{l}={ }^5 P_5=5! \\=5 \times 4 \times 3 \times 2 \times 1=120\end{array}$
(ii) If only 3 letters out of 5 are taken then the number of words
$={ }^5 P_3=\frac{5!}{2!}=\frac{120}{2}=60 .$
(iii) D comes in the beginning of every word then D becomes fixed and hence we have to arrange only 4 letters. Hence, number of words will be
$\begin{array}{l}={ }^4 P_4=4! \\=4 \times 3 \times 2 \times 1=24\end{array}$
(iv) If every word starts with D and ends with I , then D and I come fixed and hence we have to arrange only 3 letters. Hence the number of words will be
${ }^3 P_3=3!=3 \times 2 \times 1=6 $
(v) When both vowels are to be kept together then they can be considered as one letter (EI). So we will have 4 letters. These can be arranged in ${ }^4 P _4=4!=$ 24 ways i.e. (EI) also have two letters which when put together can be taken as (EI) or (IE) i.e. it is also can be arranged by $2!=2$ methods.
Thus total number of words $=2 \times 24=48$ .
(vi) If both the vowels come at the beginning of every word, then E and I (at the first two positions) become fixed and hence we have to arrange only 3 letters which are ${ }^3 P_3=3!=6$ ways. But the vowels $E$ and I can also be arranged $2!=2$ methods. In this way the total number of words will be $=2 \times 6=12$

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Question 42 Marks

Find the value of $n$ such that :
(i) ${ }^n P _5=42{ }^n P _3, n>4$
(ii) $\frac{{ }^n P _4}{{ }^{n-1} P _4}=\frac{5}{3}, n>4$

Answer

(i) Given,
$\begin{array}{ll}& { }^n P_5=42{ }^n P_3 \\\text { or } & n(n-1)(n-2)(n-3)(n-4) =42 n(n-1)(n-2)\ & \\\because & n>4 \text { therefore } n(n-1)(n-2) \neq 0\end{array}$
So, dividing both sides by $n(n-1)(n-2)$
$(n-3)(n-4)=42$
$\begin{array}{ll}\text { or } & n^2-7 n-30=0 \\ \text { or } & n^2-10 n+3 n-30=0 \\ \text { or } & (n-10)(n+3)=0 \\ \text { or } & n-10=0 \text { or } n+3=0 \\ \text { or } & n=10 \text { or } n=-3\end{array}$
Since $n$ cannot be negative, so $n=10$
(ii) Given, $\frac{{ }^n P _4}{{ }^{n-1} P _4}=\frac{5}{3}$
$\begin{aligned} \text { Similarly } 3 n(n-1)(n-2) & (n-3)=5(n-1) (n-2)(n-3)(n-4)\end{aligned}$
$ \begin{array}{l}\text {or}\quad 3 n=5(n-4) \\\quad\quad {[\text { Since }(n-1)(n-2)(n-3) \neq 0, n>4]} \end{array} $
or $\quad n=10$

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