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?

HelpBot Sep 6, 2020

#2**+2 **

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 Sep 7, 2020

#3**+1 **

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.

Guest Sep 8, 2020

#4**+2 **

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.

Melody
Sep 9, 2020

#6**+3 **

You say " Thank you very much!** But 9420 **is incorrect "!, But the answer given to you **is 9240 and NOT 9420 ?!.**

Guest Sep 11, 2020