Math, asked by ravishali0001, 1 year ago

A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 women. The shortlist consists of 9 men and 6 women. In how many ways can this be done?

Answers

Answered by qwtiger
8

Answer:

4194

Step-by-step explanation:

Since, we've to form the majority of me, therefore, there can be either 1 woman, or 2 women, or 3 at most and the rest are men in each case. For the three cases let's calculate No. of ways:

⁶C₁*⁹C₆ + ⁶C₂*⁹C₅ + ⁶C₃*⁹C₄

=6 * 84 + 15 * 126 + 20* 126 = 4194 ways

When we calculate this , we will get the desired no. of ways.

Here, The first term denotes no. of ways of selecting woman and the latter term denotes the no. of ways of selecting man.

Answered by amitnrw
5

Answer:

A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 women. The shortlist consists of 9 men and 6 women.

4914 ways

Step-by-step explanation:

to have majority of Men minimum 4 men to be there out of 7

7 members can be

4 Men  + 3 Women  = ⁹C₄⁶C₃  = 9!/(4!5!) *  6!/(3!3!)  = 2520

5 Men +  2 Women  = ⁹C₅⁶C₂  = 9!/(4!5!) *  6!/(4!2!)  =  1890

6 Men +  1 Women  = ⁹C₆⁶C₁ = 9!/(6!3!) *  6!/(5!1!)  =  504

Total number of ways = 2520 + 1890 + 504 = 4914

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