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There are 84 different combinations possible for the committee of 6, taken from

4 students and 5 teachers.

1.- The committee with 4 students has 4C4 number of combinations of 4 students out of 4 and 5C2 number of combinations of 2 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 4 students and 2 teachers.

2.- The committee with 3 students has 4C3 number of combinations of 3 students out of 4 and 5C3 number of combinations of 3 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 3 students and 6 teachers.

3.- The committee with 2 students has 4C2 number of combinations of 2 students

out of 4 and 5C4 number of combinations of 4 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 2 students and 4 teachers.

4.- The committee with 1 student has 4C1 number of combinations of 1 student out of 4 students and 5C5 number of combinations of 5 teachers out of 5 to be combined with. The product of these two give the number of different combinations possible in a committee formed by 1 student and 5 teachers.

We now add up all possible combinations:

4C4∙5C2 + 4C3∙5C3 + 4C2∙5C4 + 4C1∙5C5 = 1(10) + 4(10) +6(5) + 4(1) = 84

There are 84 different combinations possible for the committee of 6, taken from

4 students and 5 teachers.

[ nCr = n!/(r!(n-r)!) ]

[ n! = n(n-1)(n-2)∙∙∙(3)(2)(1) ]

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Q: How many ways can a committee of 6 be chosen from 5 teachers and 4 students if all are equally eligible?
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