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Question: You have 9 English books, 7 French books and 5 German books. How many ways are there to make a row of three books in which exactly one language is missing (the order of the three books makes a difference) ?

I do not want to use permutations (nPr). My solution is: $$\left(\binom{16}{3} + \binom{14}{3} + \binom{12}{3}\right)\times 3$$ My logic is: pick 3 books out of English & French (16) + pick 3 books out of English & German (14) + pick 3 books out of French & German (12) and multiply by 3 since the order matters (and there are three slots).

Is my answer correct?

Thanks!

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What is c here ? –  Thomas Apr 4 '12 at 18:51
    
@Thomas: Choice coefficient. –  Arturo Magidin Apr 4 '12 at 18:53
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There are $3!=6$ ways to arrange $3$ objects. Apart from that... –  example Apr 4 '12 at 18:54
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No, your answer is not correct. You made no distinction between picking 3 English books or the other situation 2 English and 1 French. The first situation does not work here, the second does. Try picking (1 E and 2 F), (2 E and 1 F), (1 E and 2 G), (2 E and 1 G), (1 F and 2 G), (2 F and 1 G). –  Graphth Apr 4 '12 at 18:55
    
Are there any duplicates among the books? –  JB King Apr 4 '12 at 19:26
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2 Answers

No.

If you have to have exactly two languages then you have to make sure you have both. So for English and French you need ${9 \choose 2}{7 \choose 1}+{9 \choose 1}{7 \choose 2} $ rather than ${16 \choose 3}$ and similarly with the other pairs of languages.

When you say "order matters" do you mean order of languages or order of books? If the latter then you should be multiplying by $3! = 6$.

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So: ${9 \choose 2} {7 \choose 1} + {9 \choose 1} {7 \choose 2} + {9 \choose 2} {5 \choose 1} + {9 \choose 1} {5 \choose 2} + {7 \choose 2} {5 \choose 1} + {7 \choose 1}{5 \choose 2} $ ? –  Darksky Apr 4 '12 at 19:15
    
@Nayefc: except that this doesn't take into account the order; also, it's easier to note that if you have 2 English books ($\binom{9}{2}$ ways), then you only need to select one book from the $12=7+5$ non-English books, so rather than count as $\binom{9}{2}\binom{7}{1}+\binom{9}{2}+\binom{5}{1}$, you can simply do $\binom{9}{2}+\binom{12}{1}$. To account for order, you'd have to multiply the final answer by $3!$. –  Arturo Magidin Apr 4 '12 at 19:24
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You are miscounting: $\binom{16}{3}$ tells you the number of ways of picking three books out of the 16 total books in English and French, but you have no guarantee that you will have both English and French represented: you are counting selecting only English books. Worse, that is also counted in your $\binom{14}{3}$ coefficient. So you are counting things you should not count, and counting them too many times.

There are a couple of possible approaches:

  1. Count how many ways there are of selecting $3$ books in order, and then remove the number of ways of making the selection in which all three or only one language is represented; these latter are fairly easy to count.

  2. Consider the case of exactly two English, exactly two French, or exactly two German books separately, and add the results.

Let me know if you need more help.

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Would this count it properly: ${9 \choose 2} {7 \choose 1} + {9 \choose 1} {7 \choose 2} + {9 \choose 2} {5 \choose 1} + {9 \choose 1} {5 \choose 2} + {7 \choose 2} {5 \choose 1} + {7 \choose 1}{5 \choose 2} $ ? –  Darksky Apr 4 '12 at 19:21
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See my comment above; this does not take the order of the books into account, and you can simplify matters by noting that once you select two books in one language, you just need to select a single book from the remaining off-language books. So you just need $\binom{9}{2}\binom{12}{1} + \binom{7}{2}\binom{14}{1} + \binom{5}{2}\binom{16}{1}$ (two english and one non-english; or two french and one non-french; or two german and one non-german). Then multiply by $3!$ to take order of the books on the row into account. –  Arturo Magidin Apr 4 '12 at 19:26
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