Joined: 11 Mar 2007
[, given: 0] 0
Combinatorics Question from Downloaded Materials [
03 Jun 2007, 15:48
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Could someone help me on this? It is a problem included in some of the review materials I've downloaded. The answer is 80, I'm sure of that, but I'm not confident of how I got the answer.
There are 5 married couples and a group of 3 is to be formed out of them. How many arrangements are there if a married couple may not be in the same group?
I'm thinking it's easiest to calculate all of the arrangements and then subtract out the threesomes with a married couple (get your mind out of the gutter). So here goes...
(10!)/(3!)*(7!)=120 different arrangements of the 5 married couples...
If we have a married couple seated, there are (8!)/(1!)*(7!)=8 (or since there is only 1 spot left to fill and 8 people who can fill it) different combinations for one couple.
Since there are 5 couples, 5x8=40...
So 120-40=80 possible combinations.
Thanks for your help in advance.
Joined: 04 Mar 2007
[, given: 0] 0
I have 2 other methods of calculating:
1) 1C5*2C4 + 1C5*2C4 + 3C5 + 3C5 = 80
1C5*2C4 - one man out of 5 in combination with 2 women out of 4
1C5*2C4 - one woman out of 5 in combination with 2 men out of 4
3C5 + 3C5 - 3 men out of 5 and 3 women out of 5
2) Quantity of combinations of 3 couples out of 5 couples, multiplied by 8
8*3C5 = 80
The last method is the easiest one, but I don't understand it.
Could anyone explain?