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"Two punts can each hold 6 people. A party of 10 wishes to use these punts. In how many ways can the party be divided?"

The answer is 462, but I can't for the life of me work out how to do it.

Help, please?
Reply 1
Avatar for aznkid66
aznkid66
1
You can split it into two cases, and sum the number of solutions for each case to arrive at the answer.

As a hint, the first case would be where the party is divided equally into 5 people per punt. For this case, in how many ways can the party be divided?
You also need to treat the punts as indistinguishable. Though it's not clear from the question that you need to do that.
Reply 3
Avatar for TheMan100
TheMan100
OP
9
Thanks.
I've wasted too much time on this question and I'm getting nowhere though, can somebody please just show me how to reach the answer?
Original post by TheMan100
Thanks.
I've wasted too much time on this question and I'm getting nowhere though, can somebody please just show me how to reach the answer?


2 cases:

a) 5 in each punt.

So, how many choices for the first punt? 10C5, and the 2nd punt takes whatever is left in each case.

b) 6 in one punt and 4 in the other.

Have a go.

Then add the two values.
Reply 5
Avatar for TheMan100
TheMan100
OP
9
Original post by ghostwalker
2 cases:

a) 5 in each punt.

So, how many choices for the first punt? 10C5, and the 2nd punt takes whatever is left in each case.

b) 6 in one punt and 4 in the other.

Have a go.

Then add the two values.


FINALLY :smile:

Thanks... at one point I was close, but I was doing 10C5 twice for some reason.

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