The number of ways in which we can choose a committee from four men and six women so that

Question:

The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two man and exactly twice as many women as men, is

(a) 94

(b) 126

(C) 128

(d) none of these

Solution:

Number of men = 4 

Number of women = 6    (given) 

Given condition says, committee include at-least two man and exactly twice as many women as men . 

So, we can select either 2 men and 4 women or 3 men and 6 women. 

$\therefore$ Required number of ways of forming committee is ${ }^{4} C_{2} \times{ }^{6} C_{4}+{ }^{4} C_{3} \times{ }^{6} C_{6}$

i. e $\frac{4 !}{2 ! 2 !} \times \frac{6 !}{4 ! 2 !}+\frac{4 !}{3 !} \times 1$

$=\frac{4 \times 3}{2} \times \frac{6 \times 5}{2}+4$

$=6 \times 15+4$

$=94$

Hence, the correct answer is option A.

 

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