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pickupchik [31]
3 years ago
9

How many different committees can be formed from 8 teachers and 31 students if the committee consists of 3 teachers and 3 ​stude

nts?
Mathematics
2 answers:
ivann1987 [24]3 years ago
7 0
There are many students, however, because there has to be three teachers in each committee, there can only be 2 full committees.
This is because 8 divided by three equals 2 committees, with 2 left over. Even though you have some left over, you cannot create another full committee. Hope that this helped you!

Snezhnost [94]3 years ago
6 0
I'm out at the moment, so I'll explain asap.
There are 8C3 * 31C3 ways in selecting 3 teachers and 3 students.

Since order doesn't matter, we use combinatoric, nCr, instead of permutation, nPr.

Now, we merely need to select 3 teachers from 8 teachers (objects) and 31 students (objects).

Since, we want them simultaneously, we need to multiply the two together.

In essence, there are 8C3 × 31C3 ways
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