There are 4 genres, each genre has 11 books. I want to choose 4 books such that (order does not matter):

  1. all books are from the same genre
  2. all books are from different genres
  3. the books are from exactly three genres
  4. the books are from exactly two genres

The following are my attempts:

  1. $11 \choose 4$ chooses 4 books from one genre that contains 11 books. There are 4 genres. Therefore, $4 \times {11 \choose 4}$ total combinations.

  2. $11^4$ total combinations. 11 possibilities for the first, 11 for the second, 11 for the third, 11 for the fourth. Does this take order into account? Again, I want order to be irrelevant.

  3. $11^3$ for three slots, then $10 \times 3$ for the remaining slot. $11^3 \times 30$ total combinations.

  4. ${11 \choose 2} \times {11 \choose 2} \times {4 \choose 2}$. Since you are choosing 2 books from 11 twice, and you are choosing 2 genres from 4 genres.

Any advice on these attempts would be appreciated.

  • 2
    $\begingroup$ Try number $3$ again. $\endgroup$ – saulspatz Sep 9 '19 at 23:05
  • 1
    $\begingroup$ (1) and (2) are both correct. For (3) you would notice that exactly one genre gets two books. Pick which genre it was and then pick which two books from that genre they were. Then pick the remaining two genres and which book from each it was. For (4) you were close but you only considered what happened if you pick two from each of the genres and neglected to account for what happens if you picked three from one and one from another. $\endgroup$ – JMoravitz Sep 9 '19 at 23:05
  1. Correct
  2. Correct (If you multiplied by a factor of $4!$ then order would matter, but the way you have counted says order does not matter)
  3. Not quite. Your method of counting says choose a book from each of 3 genres (but which ones?), then pick one of the genres and choose a second book from it. There are two problems with this. First, you haven't said which 3 genres are being used, so you need to multiply by a factor of $\binom{4}{3}$. Second, the order in which you choose the two books from one genre matters in your method of counting. Suppose the books are $A_1,\dots A_{11}$ for the first genre, $B_1,\dots,B_{11}$ for the second, and so on. If I choose $A_1,B_1$, and $C_1$ and then pick genre $A$ and choose $A_2$, that is counted differently from choosing $A_2$, $B_1$, and $C_1$ and then picking genre $A$ and choosing $A_1$. So you need to divide by a factor of $2$. Another way to think about this is: choose 3 genres ($\binom{4}{3}$), choose one of them to have two books ($3$), choose one book each from two genres ($11^2$), and choose two books from the third ($\binom{11}{2}$).
  4. Almost. This would be correct if the question said two books are from one genre and two books are from another. But what if 3 are from one and 1 is from another?
  • $\begingroup$ Thanks so much! So for number 4, would I take what I have already and multiply that by ${4 \choose 2} \times {11 \choose 3} \times {11 \choose 1}$? Or is this double-counting $4 \choose 2$ $\endgroup$ – theta Sep 10 '19 at 0:24
  • 1
    $\begingroup$ You don't want to multiply, you want to add because you are counting disjoint things. Also, the expression in your comment actually under counts, because you need to designate which of the genres has 3 books and which has 1. In other words, you should have $4 \cdot 3$ instead of $\binom{4}{2}$. $\endgroup$ – kccu Sep 10 '19 at 0:56

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