How many committee of size 5 consisting of 3 men and 2 women can be seated from 8 men and 6 women if a certain man must not be on the committee ?

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How many committee of size 5 consisting of 3 men and 2 women can be seated from 8 men and 6 women if a certain man must not be on the committee ?

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Answer 1 · 2017-09-21T00:04:32

There are $525$ $525$ different arrangements.

Explanation:

In the problem you have $8$ $8$ men and $6$ $6$ women.
$M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}, M_{7}, M_{8}$ $M_1, M_2, M_3, M_4, M_5, M_6, M_7, M_8$
$W_{1}, W_{2}, W_{3}, W_{4}, W_{5}, W_{6}$ $W_1, W_2, W_3, W_4, W_5, W_6$

Since one man is not allowed to be in the committee, we can take out one man from the list (it doesn't matter; for simplicity's sake, I will take out Man 8)

$M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6}, M_{7}$ $M_1, M_2, M_3, M_4, M_5, M_6, M_7$
$W_{1}, W_{2}, W_{3}, W_{4}, W_{5}, W_{6}$ $W_1, W_2, W_3, W_4, W_5, W_6$

Then we need to find out how many arrangements of men and arrangements of women, then multiply them.

Arrangements of Men
Since the order in which the men are chosen does not matter, and you cannot choose the same man twice, we can use combinations.

The formula is $n C r = \frac{n!}{r! (n - r)!}$ $nCr = (n!)/(r!(n-r)!)$ , where $n$ $n$ is the number of things to choose from and $r$ $r$ is the number of things to take out.

If we are choosing $3$ $3$ men from $7$ $7$ , we can use $n = 7$ $n = 7$ and $r = 3$ $r = 3$ :

$\frac{7!}{3! (7 - 3)!}$ $(7!)/(3!(7-3)!)$

$\frac{5040}{3! \cdot 4!}$ $(5040)/(3! * 4!)$

$\frac{5040}{6 \cdot 24}$ $(5040)/(6*24)$

$\frac{5040}{144}$ $(5040)/(144)$

$35$ $35$

So there are $35$ $35$ possible arrangements for men.

Arrangements of Women
We can do the same process for women: using combinations.

The formula is $n C r = \frac{n!}{r! (n - r)!}$ $nCr = (n!)/(r!(n-r)!)$ , where $n$ $n$ is the number of things to choose from and $r$ $r$ is the number of things to take out.

If we are choosing $2$ $2$ women from $6$ $6$ , we can use $n = 6$ $n = 6$ and $r = 2$ $r = 2$ :

$\frac{6!}{2! (6 - 2)!}$ $(6!)/(2!(6-2)!)$

$\frac{720}{2! \cdot 4!}$ $(720)/(2! * 4!)$

$\frac{720}{2 \cdot 24}$ $(720)/(2*24)$

$\frac{720}{48}$ $(720)/(48)$

$15$ $15$

So there are $15$ $15$ possible arrangements for women.

Counting Principle
Now we know that there are $35$ $35$ possible arrangements for men and $15$ $15$ possible arrangements for women. Now, according to counting principle, we can multiply $35$ $35$ and $15$ $15$ to see how many possible committee arrangements there are.

$35 \cdot 15 = 525$ $35*15 = 525$

Final answer:
There are $525$ $525$ different ways to create a committee.

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How many committee of size 5 consisting of 3 men and 2 women can be seated from 8 men and 6 women if a certain man must not be on the committee ?

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