Hello,
The easiest place to start here is with the side dishes. How many combinations of side dishes are there? For the first side dish, there are five choices. For the second side dish, there are now four choices. Since it is a combination (the order in which we pick the side dishes doesn't matter), we have to divide by 2!.
So, for the side dishes, we have (5*4) / (2*1) = 10.
Now, for each combination of side dishes (10), we have four choices of entrees.
So, 10*4= 40.
Hope this helps!
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GMat prep multiplication rule
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Stuart@KaplanGMAT
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Here's a rule you can use for both probability and combinations/permutations:
When we're asked to calculate the probability or number of MULTIPLE events occuring, we MULTIPLY the individual results.
Common words to help you identify multiple events are "and", "both" and "all".
When we're asked to calculate the probability or number of ALTERNATIVE events occuring, we ADD the individual results.
Common words to help you identify alternative events are "or", "at least" and "at most".
The question you posted says that a meal consists of 1 entree AND 2 side dishes. "And" means multiple events, so we MULTIPLY the number of possible entree selections and the number of possible side dish selections.
When we're asked to calculate the probability or number of MULTIPLE events occuring, we MULTIPLY the individual results.
Common words to help you identify multiple events are "and", "both" and "all".
When we're asked to calculate the probability or number of ALTERNATIVE events occuring, we ADD the individual results.
Common words to help you identify alternative events are "or", "at least" and "at most".
The question you posted says that a meal consists of 1 entree AND 2 side dishes. "And" means multiple events, so we MULTIPLY the number of possible entree selections and the number of possible side dish selections.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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I used the combinations formula:
C = n! / r!(n-r)!
4 Entrees:
here n = 4, r = 1 ... C1 = [4!/(1!3!)] = 4
5 side dishes:
here n = 5, r = 2 ... C2 = [5!/(2!3!)] = 10
In case of "AND" we multiply ... so C1 * C2 = 4 * 10 = 40
C = n! / r!(n-r)!
4 Entrees:
here n = 4, r = 1 ... C1 = [4!/(1!3!)] = 4
5 side dishes:
here n = 5, r = 2 ... C2 = [5!/(2!3!)] = 10
In case of "AND" we multiply ... so C1 * C2 = 4 * 10 = 40
