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Help Me

by selvaraj » Sat Mar 14, 2009 4:18 am
Dear All,

I want help form your end. Could you please help me the following questions.It is really greatfull to you.

(A) A man has 10 close friends. The man is having a party, in how many ways can he invite
(i) 5 of them?

(ii) 5 of them if two of the friends are married and will not attend separately?

(iii) 5 of them if two of them are not on speaking them and will not attend together?
Selvaraj.C
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BMA-Bangalore-560037
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Re: Help Me

by Vemuri » Sat Mar 14, 2009 8:14 am
I am giving my shot at these questions. I am sure we can learn from the gurus in this forum if what I am stating is not correct :)

1. 5 friends can be invited in 10c5 ways, i.e. 10!/(5!*5!) = 252 ways.

2. Invite 5 friends, with 2 of them married in the list of 10 friends. So, in all we have 8 friends from whom 3 friends can be selected & 2 ways in which the 2 married couple can be invited. So, 8c3+2 = 56+2 = 58 ways.

3. Since 2 of them are not in speaking terms & will not attend together, we are left with 9 friends to pick the 5, i.e. 9c5 = 126 ways.

I hope my answers are correct :oops:
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Re: Help Me

by karmayogi » Sat Mar 14, 2009 9:37 am
Vemuri wrote:I am giving my shot at these questions. I am sure we can learn from the gurus in this forum if what I am stating is not correct :)

1. 5 friends can be invited in 10c5 ways, i.e. 10!/(5!*5!) = 252 ways.

2. Invite 5 friends, with 2 of them married in the list of 10 friends. So, in all we have 8 friends from whom 3 friends can be selected & 2 ways in which the 2 married couple can be invited. So, 8c3+2 = 56+2 = 58 ways.

3. Since 2 of them are not in speaking terms & will not attend together, we are left with 9 friends to pick the 5, i.e. 9c5 = 126 ways.

I hope my answers are correct :oops:
I think, 1st is correct, while 2nd and 3rd are wrong.

My logic:
1. 5 to select out of 10
10C5 = 252
2. Out of 10, 2 are married. Hence, either both of them will be selected or none of them will be selected. There is only 1 way to select both of the them.
1(both selected)*8C3 + 8C5 (none of them selected)
56 + 56 = 112
3. Suppose A and B can't stand each other. Then, there are 3 ways to selected.
Only A is selected * 8C4 + only B is selected * 8C4 + neither A nor B selected * 8C5
=8C4 + 8C4 + 8C5
=2 * 8C4 + 8C5
=196
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Re: Help Me

by Vemuri » Sat Mar 14, 2009 10:39 pm
karmayogi wrote: I think, 1st is correct, while 2nd and 3rd are wrong.

My logic:
1. 5 to select out of 10
10C5 = 252
2. Out of 10, 2 are married. Hence, either both of them will be selected or none of them will be selected. There is only 1 way to select both of the them.
1(both selected)*8C3 + 8C5 (none of them selected)
56 + 56 = 112
3. Suppose A and B can't stand each other. Then, there are 3 ways to selected.
Only A is selected * 8C4 + only B is selected * 8C4 + neither A nor B selected * 8C5
=8C4 + 8C4 + 8C5
=2 * 8C4 + 8C5
=196
Thank you Karmayogi. I agree with your approach & answers.

What are the OAs?
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by avenus » Sun Mar 15, 2009 1:43 am
agree with karmayogi

Another way of looking at 2) 3) to reach the same results:

2)

252 - 2(8C4) = 112

ie, from the total we subtract the cases in which one of the members of the couple is in the group and the other one isn't: 2(8C4)


3)

252 - 8C3 = 196

ie, from the total we subtract the cases in which those two "friends" are part of the group at the same time: 8C3
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