(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 these 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.

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Solution

(a) (9W, 8M) committee of 12 with at least 5 women
(5W, 7M) + (6W, 6M) + (7W,5M)
+ (8W,4M)+(9W,3M)
$(^{9} C_{5} \times^{8} C_{7})$ + $(^{9} C_{6} \times^{8} C_{6})$ + $(^{9}C_{7
}\times^{8}C_{5}) $+$ (^{9}C_{8}\times^{8}C_{4}) $+$ (^{9}C_{9}\times^{8}C_{3}) $(126 x 8) + (84 x 28) + (36 x 56) + (9 x 70) + (1 x 56) I I I I I I I V V A d d I I I, I V, V f o r w o m e n t o b e i n m a j o r i t y a n d i n I m e n a r e i n m a j o r i t y . (b) 6 M, 4 F : c o m m i t t e e o f 5. p = A t l e a s t o n e l a d y = T o t a l - N o l a d y =$ ^{10}C_{5} $-$ ^{6}C_{5} $=$ \dfrac{10.9.8.7.6}{120} $- 6 = 252 - 6 = 246 q = A t l e a s t t w o m e n = T o t a l - (N o m a n) + (O n e m a n) =$ ^{10}C_{5} $- 0 + $^{6} C_{1} $ . $^{4} C_{4} $ = 252 - 6 = 246.$ \therefore $p = q o r$ \dfrac{p}{q}$ = 1.

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