# Counting problems [closed]

1) A businessman has 10 suits. He needs to pack 3 of them to go on a trip. How many can he do this?

2) A multiple choice test has 8 questions, each with 3 possible answers. how many can the test be filled?

3) I order a dozen bagels. there are poppy seed, onion and sesame bagels available today. in how many different ways can I fill my order?

4) A bag contains 26 scrabble tiles, each with a different letter on it. i draw 3 tiles and arrange them on the tray in front of me. How many "words" can I form this way?

Attempt:
1) 10 choose 3
2) 8 choose 3
3)$3^{12}$
4) $5!$

Is my thinking correct?

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## closed as off-topic by Lord_Farin, Watson, RecklessReckoner, William, LeucippusJun 13 at 0:02

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When answering questions such as these, it's best to provide an explanation as to why there's e.g. $\binom{10}{3}$ ways (rather than some other formula involving $10$ and $3$). – Douglas S. Stones Apr 15 '13 at 17:03
I'm voting to close this question as off-topic because it is multiple questions asked as one. – Lord_Farin Jun 12 at 17:01

2. Each answer can independently be chosen in 3 ways. So the total number of ways is $3^8$.
4. Three tiles could be drawn in $\binom{26}{3}$ ways, and there are $3!$ ways to permute them after drawing. So total no. of ways is $3!\binom{26}{3}$.
I don't agree with 3. You just need to count the number of solutions of $p + o + s = 12$, where $p,o,s \ge 0$ and integers. Otherwise poppy 11x + sesame is counted differently as sesame plus poppy 11x, e.g. – Henno Brandsma Apr 15 '13 at 14:24
(3) is equivalent saying how many ways can you put 12 indistinguish balls into 3 distinguish boxes. By using the the distribution rule $\binom{n+k-1}{n}$ where $n$ is the number of balls, and $k$ is the number of boxes, so, the solution is simply $\binom{12+3-1}{12}$ – user62453 Apr 15 '13 at 16:04