Question

(a) A committee of 12 is to be formed from 9 women and 8 men. In how many ways this can be done if at least five women have to be included in a committee? In how many of this committees
(i) The women are in majority?
(ii) The men are in majority?
(b) Out of 10 persons (6 males, 4 females), a committee of 5 is formed
P = number of such committees which include at least one lady
q = number of such committees which include at least two men
Find the ratio p: q

Answer 1

To solve the question given above, we will solve each part of the question separately. To solve part A we will select different number of women in different cases like 5 women and 7 men in case I, 6 women and 6 men in case II and so on. We will finally add all this cases to get the desired answer. Then, we will determine in how many of these cases, the women and men will be in majority. This will be the answer of second and third subpart of (a). To solve part B, we will first find out by selecting at least 1 women. In this also, the cases will be formed depending on the number of women selected. Similarly, we will find q and finally take their ratio.

Complete step-by-step answer:
Here, in this question, we have two parts and in these two parts also, we have subparts. So we will solve each part separately.
(a) In this part, we are given that a committee of 12 is to be formed but the condition given is that no matter how committee is formed, there should be at least 5 women in each case, so, we can say that the number of women in committee can be 5, 6, 7, 8 or 9. Thus, we are going to form different cases according to this. Thus we have:
Case I: The committee has 5 women. If there are 5 women then there are 7 men in a committee.

& \text{The number of ways of selecting 5 women}={}^{9}{{C}_{5}} \\\ & \text{The number of ways of selecting 7 men}={}^{8}{{C}_{7}} \\\ & \text{Thus total number of ways}={}^{9}{{C}_{5}}\times {}^{8}{{C}_{7}} \\\ \end{aligned}$$ Case II: The committee has 6 women. Now the number of men will be = 6. Similarly from first case, the number of ways of selecting them in a committee $$\Rightarrow {}^{9}{{C}_{6}}\times {}^{8}{{C}_{6}}$$ Case III: The committee has7 women. Now the number of men will be 5. The number of ways of selecting them in a committee $$\Rightarrow {}^{9}{{C}_{7}}\times {}^{8}{{C}_{5}}$$ Case IV: The committee has 8 women. Now the number of men will be =4. The number of ways of selecting them in a committee $$\Rightarrow {}^{9}{{C}_{8}}\times {}^{8}{{C}_{4}}$$ Case V: The committee has 9 women. Now the number of men will be = 3. The number of ways of selecting them in a committee $$\Rightarrow {}^{9}{{C}_{7}}\times {}^{8}{{C}_{3}}$$ Thus, the total number of ways can be obtained by adding all the cases. On adding, we get: $$\begin{aligned} & \text{Total ways}=\left( {}^{9}{{C}_{5}}\times {}^{8}{{\text{C}}_{7}} \right)+\left( {}^{9}{{C}_{6}}\times {}^{8}{{\text{C}}_{6}} \right)+\left( {}^{9}{{C}_{7}}\times {}^{8}{{\text{C}}_{5}} \right)+\left( {}^{9}{{C}_{8}}\times {}^{8}{{\text{C}}_{4}} \right)+\left( {}^{9}{{C}_{9}}\times {}^{8}{{\text{C}}_{3}} \right) \\\ & \Rightarrow \left( 126\times 8 \right)+\left( 84\times 28 \right)+\left( 36\times 56 \right)+\left( 9\times 70 \right)+\left( 1\times 56 \right) \\\ & \Rightarrow \text{1008 + 2352 + 2016 + 630 + 56} \\\ & \Rightarrow \text{6062} \\\ \end{aligned}$$ Thus, there are 6062 ways of making a committee in which at least there are 5 women. (i) In this part, we have to determine which the cases are when women are in majority. From above cases, case III, case IV and case V satisfy our condition. Thus, the number of committees where women are in majority is obtained by adding all the three cases. Therefore, total cases $$\begin{aligned} & \Rightarrow \text{case III}+\text{ case IV}+\text{ case V} \\\ & \Rightarrow \text{2}0\text{16 + 63}0+\text{56} \\\ & \Rightarrow \text{27}0\text{2} \\\ \end{aligned}$$ (ii) There is only one case i.e. case I where men are in majority. Thus, the number of committees where men are in majority = 100%. (b) In this question, first we will calculate P and q by forming cases according to information given In question and then we will find their ratio Calculation of P: We have to find the number of committee which have at least one lady. Thus the committee can have number of ladies = 1, 2, 3 or 4. Now, we will form the cases: Case I: When there is one lady the number of males will be 4. The number of ways of doing this are $$\Rightarrow {}^{4}{{\text{C}}_{1}}\times {}^{6}{{\text{C}}_{4}}$$ Case II: When there are two females and 3 males, the number of committees will b $$\Rightarrow {}^{4}{{\text{C}}_{2}}\times {}^{6}{{\text{C}}_{3}}$$ Case III: When there are three females and 2 males, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{3}}\times {}^{6}{{\text{C}}_{2}}$$. Case IV: When there are four females and 1 male, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{4}}\times {}^{6}{{\text{C}}_{1}}$$ Thus, $$\begin{aligned} & \text{P = Case I + Case II + Case III + Case IV + Case V} \\\ & P=\left( {}^{4}{{C}_{1}}\times {}^{6}{{\text{C}}_{4}} \right)+\left( {}^{4}{{C}_{2}}\times {}^{6}{{\text{C}}_{3}} \right)+\left( {}^{4}{{C}_{3}}\times {}^{6}{{\text{C}}_{2}} \right)+\left( {}^{4}{{C}_{4}}\times {}^{6}{{\text{C}}_{1}} \right) \\\ & \text{P=}\left( 4\times \text{15} \right)+\left( 6\times \text{20} \right)\text{+}\left( 4\times \text{15} \right)\text{+}\left( 1\times \text{6} \right) \\\ & \text{P = 60 +120 + 60 + 6} \\\ & \text{P = 246} \\\ \end{aligned}$$ Calculation of Q: We haven find the number which includes at least two men. Thus the committee can have number of men =1, 2, 3, 4, 5. Now, we will form the cases: Case l: When there are 2 males and 3 females, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{3}}\times {}^{6}{{\text{C}}_{2}}$$ Case II: When there are 3 males and 2 females, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{2}}\times {}^{6}{{\text{C}}_{3}}$$ Case III: When there are 4 males and 1 female, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{1}}\times {}^{6}{{\text{C}}_{4}}$$ Case IV: When there are 5 males and no female, the number of committees $$\Rightarrow {}^{4}{{\text{C}}_{0}}\times {}^{6}{{\text{C}}_{5}}$$ Thus, $$\begin{aligned} & \text{q = Case I + Case II + Case III + Case IV} \\\ & \text{q}=\left( {}^{4}{{C}_{3}}\times {}^{6}{{\text{C}}_{2}} \right)+\left( {}^{4}{{C}_{2}}\times {}^{6}{{\text{C}}_{3}} \right)+\left( {}^{4}{{C}_{1}}\times {}^{6}{{\text{C}}_{4}} \right)+\left( {}^{4}{{C}_{0}}\times {}^{6}{{\text{C}}_{5}} \right) \\\ & \text{q =}\left( 4\times \text{15} \right)+\left( 6\times \text{20} \right)+\left( 4\times \text{15} \right)+\left( 1\times \text{6} \right) \\\ & \text{q = 60 + 120 + 60 + 6} \\\ & \text{q = 246} \\\ \end{aligned}$$ Therefore, the ratio $$\text{P:q =}\dfrac{P}{q}$$ $$\begin{aligned} & \text{P:q =}\dfrac{246}{246} \\\ & \text{P:q =1:1} \\\ \end{aligned}$$ **Note:** In the part (b), in the calculation of q, instead of forming all the cases, we could have just added the one case (in which there are 5 males) and subtracting other case (in which there is 1male) from the cases of calculation of P.