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? 28 ways 36 ways 56 ways 84 ways

Answer 1

The number of ways the first group of three wolves (necessary to include the male wolf) is given by: Option A: 28

What is the rule of product in combinatorics?

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in $p \times q$ ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

For the considered case, we have:

• There are 3 wolves to be included in first group.
• There are 8 female wolves, and 1 male wolf.
• 1 male wolf is sure to be in the first group.

So 2 spaces are remaining and are to be filled by 2 of 8 wolves.

Assuming that all 8 of them are distinguishable, choosing 2 of them is done in $^8C_2 = 28$ ways.

The first process of choosing male wolf for first place is done in 1 way only.

And the second process of choosing 2 female wolves can be done in 28 ways(ordering doesn't matter),

and since first and second process have no necessary order, using product rule, we get;

Number of ways of forming first group of wolves = $1 \times 28 = 28$ ways.

Thus, the number of ways the first group of three wolves (necessary to include the male wolf) is given by: Option A: 28

Learn more about product rule here:

https://brainly.com/question/2763785

Answer 2

Answer:

The above answer is correct! The right answer is option A. 28 :)

Hope this helped! Brainliest would be greatly appreciated :D

Step-by-step explanation: