Question

Nine wolves, eight female and one male, are to be released into the wild three at a time. If the male wolf is to be in the first released group and order does not matter, in how many ways can the first group of three wolves be formed?

Answer 1

$\displaystyle \binom{8}{2}=\dfrac{8!}{2!6!}=\dfrac{7\cdot8}{2}=28$

Answer 2

We have been given that Nine wolves, eight female and one male, are to be released into the wild three at a time.

We need to choose 1 male wolf out of 1 wolf male. So  we can choose one wolf as:

$_{1}^{1}\textrm{C}$

$\frac{1!}{1!(1-1)!}$

$\frac{1!}{1!\cdot 0!}$

$\frac{1}{1\cdot 1}=1$

We can choose 1 male wolf in only 1 way.

Since the female wolfs are identical (order doesn't matter), so we will use combinations.

We can choose 2 female wolves out of 8 as:

$_{2}^{8}\textrm{C}$

$\frac{8!}{2!(8-2)!}$  

$\frac{8\cdot 7\cdot 6!}{2\cdot 1\cdot 6!}=\frac{8\cdot7}{2}=28$

Therefore, we can choose 2 female wolves out of 8 female wolves in 28 ways.

To find number of ways in which the first group of three wolves can be formed we will multiply the ways of choosing 1 male wolf and 2 female wolves.

$\text{Number of ways of forming first group of three wolves}=1\times 28=28$

Therefore, the first group of three wolves can be formed in 28 ways.