Help Me

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

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

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

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?

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