Question

suppose that a department contains 11 men and 17 women. how many different committees of 6 members are possible if the committee must have strictly more women than men?

Answer 1

If a department has 11 male employees and 17 female employees, then 198968 different committees of 6 members are possible with the condition that the committee has strictly more female employees than male.

As per the question statement, a department has 11 male employees and 17 female employees.

We re required to calculate the total number different committees that can be formed with 6 members strictly having more female employees than male.

Now for committee of 6 members to have more women than men, there can be two combinations:

(4 women and 2 men)

Or, (5 Women and 1 man).

That is, we will need to calculate the number of combinations we can have by selecting 4 female employees from a group of 17 and 2 Male employees from a group of 11 and the number of combinations we can have by selecting 5 female employees from a group of 17 and 1 Male employee from a group of 11 and add up the two number of combinations to obtain our required answer.

Then comes the most important thing to know to be able to solve this question, i.e., the formula to calculate combinations, which goes as

$nCr=\frac{n!}{r!(n-r)!}$

Therefore, the total number different committees that can be formed with 6 members strictly having more female employees than male is

$[(17C4)*(11C2)+(17C5)*(11C1)]\\=[(2380*55)+(6188*11)]\\=(130900+68068)\\=198968$

• combination(s): In mathematics, a combination is a way of selecting items from a collection or set, where the order of selection does not matter, i.e., for example, we have a set of three numbers X, Y and Z and then, in how many ways can we select two numbers from each set, is defined by combination.

To learn more about Combinations, click on the link below.brainly.com/question/8044761

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