| Quote: | | | |
| There are 13 entertainers. 8 are singers. 5 are comedians. A concert is to be put up by 5 of them.
There must be at least 1 comedian, but the number of singers must always be more than the number of comedians.
In how many ways can the 5 entertainers be chosen? | |
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BlackBlackCat is half-correct.
This is not probabilities (emaths), but permutation and combination (amaths).
we start by listing the possible combinations of entertainers.
As BlackBlackCat as said... there are 2 possible ways to combine them.
scenario 1 - 1 out of 5 possible comedians, 4 out of 8 possible singers.
scenario 2 - 2 out of 5 possible comedians, 3 out of 8 possible singers.
now we need to combine them one more time them because in the first scenario, that one comedian can be any of the 5 selectable comedians.
thus the number of ways to chose the 5 entertainers is as follows.
answer = (number of possible scenario 1) + (number of possible scenario 2)
we need to split them up to calculate.
for scenario 1:
5C1 X 8C4 = 350
(notice your scientific calculator has the nCr button)
for scenario 2:
5C2 X 8C3 = 560
answer = 350 + 560 = 910
Therefore, out of the 8 singers and 5 comedians, the 5 entertainers can be chosen in 910 different ways.
