another permutation combination problem

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desihokie: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jan 13, 2008 10:47 pm; Thanked: 1 times

by desihokie » Fri Jan 18, 2008 12:14 am

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Source: Beat The GMAT — Problem Solving | Fri Jan 18, 2008 12:14 am

Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

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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desihokie: Senior | Next Rank: 100 Posts; Posts: 49; Joined: Sun Jan 13, 2008 10:47 pm; Thanked: 1 times

by desihokie » Fri Jan 18, 2008 8:43 am

Thanks, stuart!

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preciousrain7: Senior | Next Rank: 100 Posts; Posts: 78; Joined: Thu Dec 06, 2007 4:16 pm; Location: NYC, NY; Thanked: 2 times

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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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pmFollowed by: 614 members; Location: Toronto; Thanked: 1710 times; GMAT Score:800

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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