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another permutation combination problem
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Stuart@KaplanGMAT
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If we can only have one person from each couple, and we have to choose a person from 3 of the 4 couples, then we start with 4C3 = 4/1 = 4
However, from each couple we're choosing 1 of the 2 people. So, we have:
4C3 * 2C1 * 2C1 * 2C1 = 4 * 2 * 2 * 2 = 32 possible teams of 3.
However, from each couple we're choosing 1 of the 2 people. So, we have:
4C3 * 2C1 * 2C1 * 2C1 = 4 * 2 * 2 * 2 = 32 possible teams of 3.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
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preciousrain7
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Why is it only 3 2C1 and not 4 2C1 since we have to choose from 4 couples? THANKS A BUNCH!Stuart Kovinsky wrote:If we can only have one person from each couple, and we have to choose a person from 3 of the 4 couples, then we start with 4C3 = 4/1 = 4
However, from each couple we're choosing 1 of the 2 people. So, we have:
4C3 * 2C1 * 2C1 * 2C1 = 4 * 2 * 2 * 2 = 32 possible teams of 3.
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Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
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- Location: Toronto
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The 2C1 refers to the 3 couples from whom we're choosing a member. Since we're only choosing 3 people, we only do the 2C1 three times.preciousrain7 wrote:Why is it only 3 2C1 and not 4 2C1 since we have to choose from 4 couples? THANKS A BUNCH!Stuart Kovinsky wrote:If we can only have one person from each couple, and we have to choose a person from 3 of the 4 couples, then we start with 4C3 = 4/1 = 4
However, from each couple we're choosing 1 of the 2 people. So, we have:
4C3 * 2C1 * 2C1 * 2C1 = 4 * 2 * 2 * 2 = 32 possible teams of 3.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
