Different arrangements - Permutation

Question

Into how many different arrangements that look different can three identical trigonometry books, 4 identical calculus books, 5 identical algebra books be placed on a shelf?

I assume the word "books" are meant to be placed after each branch of mathematics? — MT_, Mar 03 '14 at 00:23
We are looking at the $12$-letter words made up of the letters t,t,t,c,c,c,c,a,a,a,a,a. — André Nicolas, Mar 03 '14 at 00:32

score 1 · Answer 1 · answered Mar 03 '14 at 00:25

1

Start by simply counting how many ways there are to shelve the 12 books. Then modify the number you get to account for fact that you could swap around, say, the trigonometry books with one another and still get the same arrangement.

answered Mar 03 '14 at 00:25

Nick

1,670

score 0 · Answer 2 · answered Mar 03 '14 at 00:25

0

(3+4+5)!/(3!4!5!) = 12!/(3!4!5!) = 27,720 ways.

answered Mar 03 '14 at 00:25

DeepSea

77,651

score 0 · Answer 3 · answered Mar 03 '14 at 00:35

We are looking at the $12$-letter words made up of the letters t,t,t,c,c,c,c,a,a,a,a,a.

We have $12$ "slots." The slots for the t's can be chosen in $\binom{12}{3}$ ways. For each choice, the slots for the c's can be chosen in $\binom{9}{4}$ ways. And once we have done this, there is only $1$ way to place the a's (or if you prefer there are $\binom{5}{5}$ ways).

Different arrangements - Permutation

3 Answers3