I think I've got the basics of combinatorics down.
I just don't know when to add possibilites and when to multiply.
The fundamental counting principal says to multiply - but can someone outline an instance of where you would add up two permutations and a combination as opposed to just multiplying all 3 of the quotients?
I know to multiply when the events are separate and independent of each other but the MGMAT study guides don't have an instance where you would add but I'm sure I've run into an example of adding somewhere but I just forgot to write it down.
Thanks
Combinatorics - when to add when to multiply
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You multiply when you take a sequence of actions to get the desired outcome.
You add when you break the outcome into different cases. You add the cases to get the total.
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed?
Sequence of actions: choose the men, then choose the women
7C4 * 5C2
Here you multiply.
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed if Jenny and John refuse to be on the committee together?
Case 1: Both Jenny and John are not on the committee.
Sequence of actions:Choose 4 men from the remaining 6 men; choose 2 women from the remaining 4 women
6C4 * 4C2
Case 2: Jenny is on the committee, and John is not.
6C4 * 4C1
Case 3: John is on the committee, and Jenny is not.
6C3 * 4C2
Total: 6C4 * 4C2 + 6C4 * 4C1 + 6C3 * 4C2
You add when you break the outcome into different cases. You add the cases to get the total.
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed?
Sequence of actions: choose the men, then choose the women
7C4 * 5C2
Here you multiply.
From a group of 7 men and 5 women, how many different committees consisting of 4 men and 2 women can be formed if Jenny and John refuse to be on the committee together?
Case 1: Both Jenny and John are not on the committee.
Sequence of actions:Choose 4 men from the remaining 6 men; choose 2 women from the remaining 4 women
6C4 * 4C2
Case 2: Jenny is on the committee, and John is not.
6C4 * 4C1
Case 3: John is on the committee, and Jenny is not.
6C3 * 4C2
Total: 6C4 * 4C2 + 6C4 * 4C1 + 6C3 * 4C2
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