# Determine the number of ways n distinct marbles can be placed inside five jars.

Determine the number of ways n marbles can be placed inside five distint jars, if the 1st jar must contain 1 marble, 2nd jar must contain 4 marbles, 3rd jar must contain 5 marbles, 4th and 5th marble must contain 1 marble.

I figured this would mean subset is ($$a_1$$, $$a_2$$ + 4, $$a_3$$ + 9, $$a_4$$ + 9, $$a_5$$ + 9) But I am not sure how to approach after this. Thank you for the help.

I know that base case for n is 12 marbles which will provide 1 combination does that mean the answer is $$\binom{n-7}{5}$$

This somehow does not seem right

• I am "sure ?" how to approach this? Jun 16, 2019 at 16:39
• Are the jars distinguishable or indistinguishable? Jun 16, 2019 at 16:43

Straightforward use of binomial theorem. $$n\binom{n-1}{4}\binom{n-5}{5}(n-10)(n-11)$$ I am assuming $$n\ge 12$$ and the order of the the jars doesn't matter.
You're trying to solve the equation $$a_1+a_2+a_3+a_4+a_5+12 = n$$
So we have 5 categories and n-12 objects. So we can use the "stars and bars" method, sometimes known as combinations with repetition. We have to place 4 dividers amongst the n-12 objects. In other words, we have n-12+4=n-8 positions and want to choose 4 positions to place the dividers. The number of slots between divider $$i$$ and $$i+1$$ will be the value of $$a_{i+1}$$, with the proper understanding that the slots before the first divider is the value for $$a_1$$ and after the 4th divider constitutes $$a_5$$.
Thus we have $$\binom{n-8}{4}=\binom{n-8}{n-12}$$ ways to solve the equation.