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Intern
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Combinations [#permalink]  08 Sep 2004, 09:40
Hi,

Could sb in detail provide me clear and detail explananation of the following combination problems. I am bit confused about these.

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

a) Where 2 of the friends are married and will not attend separately.

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

Thanks.
Joined: 31 Dec 1969
Location: India
Concentration: General Management, Strategy
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[#permalink]  08 Sep 2004, 10:23
Could sb in detail provide me clear and detail explananation of the following combination problems. I am bit confused about these.

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

a) Where 2 of the friends are married and will not attend separately.

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

this is a conbination with a restriction.
The couples must always be on the same team. In that case Fix the position of the couples first. That leaves 9 places with 3 people and that becomes
9C3 = 84

With the second one, the two friends will not attend together.
Find the total number of ways without restriction and subtract the number so ways in which the tow will come together
Total number of ways without restriction is 11C5 = 462.
Number of ways in which both member will be on the team = 9C3 = 84

Required probability = 462-84 = 378
Director
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[#permalink]  08 Sep 2004, 10:25
a) The woman has a choice to call the couple in 2C2 ways. She can call the 'other' three friends out of the 9 people left in 9C3 ways.

Alternately the woman can call all the 5 friends from the pool of 9 friends in 9C5 ways.

So, we have 9C3*2C2 + 9C5 = 210 ways

b) Off the two friends who are now foes, she can call only one of them in 2C1 ways. The woman can call 4 of the remaining 9 friends in 9C4 ways.

So we have 9C4 * 2C1 = 252 ways.

Last edited by venksune on 08 Sep 2004, 11:02, edited 1 time in total.
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Re: Combinations [#permalink]  08 Sep 2004, 10:31
sair wrote:
Hi,

Could sb in detail provide me clear and detail explananation of the following combination problems. I am bit confused about these.

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

a) Where 2 of the friends are married and will not attend separately.

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

Thanks.

let me try:

a) Let AB represent the couple who won't attend seperately. Two cases:

Case 1- AB invited.
So number of selections possible - 9C3*2C2 = 84

Case 2 - AB not invited
So now number of selections possible - 9C5 = 126

So total selections for a) = 126+84 = 210

b) PQ represents the not-speaking friends. Three cases

Case 1 - P comes and so Q does not come
Number of selections = 1C1*9C4 = 126

Case 2 - Q comes and P does not come
As above, selections = 126

Case 3 - P and Q, both are not invited
Number of selections = 9C5 = 126

So total number of selections for b) = 2*126 + 126 = 378

What is the OA?
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[#permalink]  08 Sep 2004, 12:44
Only Venksune has given the right answers to both a and b.

Answers.

a) 210
b) 252
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[#permalink]  08 Sep 2004, 18:50
venksune wrote:
a) The woman has a choice to call the couple in 2C2 ways. She can call the 'other' three friends out of the 9 people left in 9C3 ways.

Alternately the woman can call all the 5 friends from the pool of 9 friends in 9C5 ways.

So, we have 9C3*2C2 + 9C5 = 210 ways

b) Off the two friends who are now foes, she can call only one of them in 2C1 ways. The woman can call 4 of the remaining 9 friends in 9C4 ways.

So we have 9C4 * 2C1 = 252 ways.

venksune - in case b, why should we not consider the possibility that the woman does not invite either one of the foe/friends - we did consider this possibility in case 1.

I am missing something here
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[#permalink]  08 Sep 2004, 19:08
stuti wrote:
venksune - in case b, why should we not consider the possibility that the woman does not invite either one of the foe/friends - we did consider this possibility in case 1.
I am missing something here

1. I am still trying to figure this out. My answer is "The third case is inturn covered in the other possibilities", but I am trying to prove this.

2. Another way to think of it:
Total number of ways (no restriction) = 11C5 = 462
If you notice, the question a and b are complement of each other.
Since ans for a = 210, b should be 252.
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[#permalink]  08 Sep 2004, 19:15
stuti wrote:
venksune - in case b, why should we not consider the possibility that the woman does not invite either one of the foe/friends - we did consider this possibility in case 1.
I am missing something here

I too selected a.210 b.252, but I think we have to consider two-guys-wont-attend in both questions. It should be a.210 and b.378

Last edited by hardworker_indian on 08 Sep 2004, 19:15, edited 1 time in total.
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[#permalink]  08 Sep 2004, 19:15
I agree with you stuti. The problem is little ambiguous and Iam not convinced that my presumption is right. I read b) little differently. 'Will not attend together' - that one of the two will attend for sure. Else, 9C5 should be added and the result will be 378. In a) either both of them attend or both dont attend.
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[#permalink]  08 Sep 2004, 19:17
1. A woman has 11 close friends.
Find the number of ways she can invite 5 of them to dinner :

a) Where 2 of the friends are married and will not attend separately.

We have 2 friends who are married and the must attend together. So now we have to pick another 3 people out of 9 to join these two friends, this gives us 9C3 = 84

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

The number of possible combinations of picking 5 people out of 11 for a dinner is 11C5 = 462
We have 2 friends who are not on speaking terms, so they must not attend together. Assuming these 2 friends are in a same group, then that group can be matched to 9C3 combinations of remaining friends = 84
So possible number of combinations = 462-84= 378
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[#permalink]  09 Sep 2004, 06:14
What is wrong with ywilfred's answer. I also did that way.
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[#permalink]  09 Sep 2004, 06:27
venksune wrote:
a) The woman has a choice to call the couple in 2C2 ways. She can call the 'other' three friends out of the 9 people left in 9C3 ways.

Alternately the woman can call all the 5 friends from the pool of 9 friends in 9C5 ways.

So, we have 9C3*2C2 + 9C5 = 210 ways

b) Off the two friends who are now foes, she can call only one of them in 2C1 ways. The woman can call 4 of the remaining 9 friends in 9C4 ways.

So we have 9C4 * 2C1 = 252 ways.

For the first why can't it be 11C5 - 9C3*2C2
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[#permalink]  09 Sep 2004, 07:02
Complement principle (Total possibilities - other possibilities = Only posibilities) approach is best suited when other possibilities are given. Else best to stick to the approach I have outlined. Also, your approach has not taken into consideration both of them not attending.
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[#permalink]  09 Sep 2004, 09:33
Hi Folks,

Just joined the group today. Want to know why didn't we consider 11C2 for the two friends who r married because we don't know among 11 friends that who is married to who ? So we can count on all possible combinations of persons married to one another than we can use 9C3 for the remaining 3 persons to be selected.

Anybody?
[#permalink] 09 Sep 2004, 09:33
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