Out of the men the number of ways you can choose 2 is 10C2 = (10*9) / (2*1)
= 45 ways
out of the women number of ways = 10C4 = (10*9*8*7) / (4*3*2*1)
= 210 ways
So the number of committees possible = 45 * 210 = 9450.
The solution of the given problem is determined by using concept of combination. Here, the answer is 9450 committees are possible.
Given :
A certain club contains 10 men and 10 women. a six-member committee consisting of 2 men and 4 women must be chosen from the club.
Find :
Number of committees are possible.
Now, solve it by concepts of combination.
According to question , there are 10 men but for committee only 2 men require. So, the possible combination is [tex]10_C_2[/tex] that is calculated below.
[tex]\begin{aligned}^{10}C_2&=\dfrac{10!}{(2!)(10-2)!}\\ &= \dfrac{10\times9\times8!}{(2!)(8!)}\\ &= 5\times9\\&=\bold{45\;ways}\end{aligned}[/tex]
There are 10 women but for committee only 4 women require. So, the possible combination is [tex]10_C_4[/tex] that is calculated below.
[tex]\begin{aligned}^{10}C_4&=\dfrac{10!}{(4!)(10-4)!}\\ &= \dfrac{10\times9\times8\times7\times6!}{(4!)(6!)}\\ &= 5\times3\times2\times7\\&=\bold{210\;ways}\end{aligned}[/tex]
Now, multiplying the both possible ways of women and men,
[tex]i.e. 45\times210 =\bold{9450\:ways}[/tex].
There are total 9450 committees are possible.
For further information, refer to this link below:
https://brainly.com/question/15301090
