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# Combinations problem

Author Message
Intern
Joined: 27 Dec 2004
Posts: 6

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17 Mar 2005, 13:08
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Can anyone help me with this problem?

A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner if

a.) 2 of the friends are married and will not attend seperately?

Ans: 210

b.) 2 of the friends are not on speaking terms and will not attend together?

Ans: 252

Thanks for the assistance
Senior Manager
Joined: 15 Mar 2005
Posts: 418
Location: Phoenix

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17 Mar 2005, 14:46
Lets take them one by one.

[I am assuming that 2 people (say A and B) are married and both are a part of the 11]

Case when the couple (group of two) attends + the case when the couple doesn't attend

When the couple attends, you need to choose 3 out of remaining 9. C(9,3)
When they don't, you need to choose 5 out of remaining 9. C(9, 5).

The answer therefore = C(9, 3) + C(9,5) = 84 + 126 = 210.

b) The total ways of inviting = C(11, 5) = 462.

The ways in which these 2 wud not be invited together = total ways of invitation - ways in which they would be invited together.

Now the ways in which they're invited together:
Assume they make 1 group. So ways in which they're invited together = C (9, 3) + C(9, 5) // when they're invited + when they're not invited
= 210.

Therefore, the ways in which they would not be invited together
= 462 - 210 = 252.

Note that the two being invited together is the same as the "invited together" case of the couple in (a).

Hope this helps.
_________________

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Manager
Joined: 01 Feb 2005
Posts: 63
Location: NYC

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17 Mar 2005, 15:51
nice explanations, kapslock
Intern
Joined: 27 Dec 2004
Posts: 6

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18 Mar 2005, 08:27
Thanks for your assistance. This site is great!!!!!
18 Mar 2005, 08:27
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