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There are 30 kids and 6 parents. Two groups are formed, one of 10 kids and 2 parents,...

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math30fail | Student, Undergraduate | Honors

Posted March 16, 2012 at 3:14 AM via web

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There are 30 kids and 6 parents. Two groups are formed, one of 10 kids and 2 parents, the other of 20 kids and 4 parents. How many different  groups of 10 kids can be formed?

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted March 16, 2012 at 3:39 AM (Answer #1)

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The key here is to determine whether we're counting possible combinations of parents in that group of 10 kids. Let's calculate based on both situations.

Let's start by assuming we're not considering parents. To find the number of possible combinations of 10 kids, we can think of it intuitively. We have 10 possible spots and 30 kids to fill them. The first spot has 30 kids to choose from, the next spot has 29 kids to choose from, and so on in the following way:

`C' = 30*29*28*27*26*25*23*22*21 = (30!)/(20!)`

However, there is a problem here, we are treating the group as if order matters in the selection of children in a group. However, if you have ever tried to deal with a group of children, you would know that order will never matter with a group of children! Therefore, we have overcounted by a factor of the number of ways we can order each group's position, which is given by 10!:

`C = C'/(10!) = (30!)/(20!10!)`

You might notice that this formula is equivalent to `((30),(10))`. This combination function is designed specifically for this sort of situation, where we are selecting a group in which the order of selection does not matter. In other words, when we are finding possible "combinations," we use the "combination" function!

All that's left now is to perform the calculation:

`((30),(10)) = (30!)/(20!10!) = 30045015`

Therefore, there are 30045015 combinations of 10 kids that can be made from 30.

Now, if we start to consider possible combinations of adults, then we combine two combinations. First, we are finding the number of possible combinations of 10 kids from a set of 30 kids. Then, we multiply that by the number of possible combinations of 2 adults from a set of 6:

`C = ((30),(10))((6),(2)) = (30!)/(20!10!)*(6!)/(4!2!)=450675225`

Therefore, if we take into account the multiple combinations of parents, we have the larger 450675225 possible combinations to make the 10-kid group.

I hope this helps!

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