+479 How many ways are there to divide 12 people into a group of 3, a group of 4, and a group of 5, if Henry has to be in the group of 4?
+118608 I am not sure but this is my best guess.
Forget about Henry, there are 11 people which are to be divided into groups of 3,3, and 5
There are 11C5 ways to choose the group of 5
Now there are 6C3 ways to choose a group of 3 BUT this is double counting, I need to divide by 2
so I have 11C5 * (6C3 /2)
Now there are 2 options for where to put Henry.
so I now have 11C5 * (6C3 /2) * 2 = 462*20/2*2 = 9240 option.
Melody's solution is accurate. It is the same as: 11! /(5! * 3! * 3!) =9,240 ways. Henry can sit with any group of 3.
+118608 ok, thanks But I would prefer to describe it as
11! /(5! * 3! * 3!) divide by 2 because the 3! and 3! are interchangeable. multiply by 2 because there are 2 ways to add in Henry.
= 9,240 ways
I am still having some trouble thinking about it this way though.
This would normally be the answer to a question like this.
How many ways can the 11 letters aaabbbcccc be arranged in a line. 11!/(3!3!4!) is the answer. [I am not going to concern myself with Henry]
For our question it would be how many ways can 11 objects be put in a line if the order of the first 5 does not matte, the order of the next 3 does not matter and the order ot the last 3 does not matter AND the second and third groups of 3 can be swapped around.
NOW it does make sense to me. 11!/(3!3!4!) divided by 2
Thanks guest for making me think about that.
You say " Thank you very much! But 9420 is incorrect "!, But the answer given to you is 9240 and NOT 9420 ?!.
