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another permutation combination problem

Expert replies
by desihokie » Fri Jan 18, 2008 12:14 am
powerprep Q
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Source: — Problem Solving |

by Stuart@KaplanGMAT » Fri Jan 18, 2008 1:36 am
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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by desihokie » Fri Jan 18, 2008 8:43 am
Thanks, stuart!
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by preciousrain7 » Sat Jan 19, 2008 10:02 am
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.
Why is it only 3 2C1 and not 4 2C1 since we have to choose from 4 couples? THANKS A BUNCH!
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by Stuart@KaplanGMAT » Sat Jan 19, 2008 10:58 am
preciousrain7 wrote:
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.
Why is it only 3 2C1 and not 4 2C1 since we have to choose from 4 couples? THANKS A BUNCH!
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.
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