Statistics and Maple: Constructing a Team of 4 from 3 Groups"

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SUMMARY

The discussion focuses on constructing a team of 4 individuals from three groups containing 21, 18, and 50 members, respectively. The total number of combinations without restrictions is calculated using Maple's numbcomb(89, 4). However, to ensure at least one member from each group, the solution involves selecting one person from each group first, creating a partial team of 3, and then selecting an additional member from the remaining pool. This method avoids the complexity of subtracting illegitimate combinations.

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Mathman2013
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Homework Statement



We have three groups, group 1 contains 21 people, group two contains 18 people and group 3 contain 50 people.

First we need to construct a team of 4 people of three groups.

How many ways can such a team be constructed?

I use combinate such that it will be calculated in Maple.

numbcomb(21+18+50, 4) is number of ways this group can constructed.


second question is tricky one.

How many ways can the group of 4 be constructed if its a requirement that there must be at least one person from each group in the group of 4 ?

The Attempt at a Solution



I know its numbcomb(21+18+50,4) to be used to again. But I am not sure howto proceed.

So any hints of help would be apriacated.
 
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Mathman2013 said:

Homework Statement



We have three groups, group 1 contains 21 people, group two contains 18 people and group 3 contain 50 people.

First we need to construct a team of 4 people of three groups.

How many ways can such a team be constructed?

I use combinate such that it will be calculated in Maple.

numbcomb(21+18+50, 4) is number of ways this group can constructed.second question is tricky one.

How many ways can the group of 4 be constructed if its a requirement that there must be at least one person from each group in the group of 4 ?

The Attempt at a Solution



I know its numbcomb(21+18+50,4) to be used to again. But I am not sure howto proceed.

So any hints of help would be apriacated.

Using numbcomb(89,4) again would be a mistake, unless you were willing to perform the very laborious process of then subtracting out the "illegitimate" combinations.

Much easier: first make sure you choose 1 person from each group, giving you a partial group of 3 people. Then you need to choose 1 more person.
 

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