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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?

1) 10 choose 3
2) 8 choose 3
4) $5!$

Is my thinking correct?

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – RecklessReckoner, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.

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
  1. Your answer is correct.
  2. Each answer can independently be chosen in 3 ways. So the total number of ways is $3^8$.
  3. Your answer is incorrect. (See the comment below.)
  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 $3!\binom{26}{3}$.
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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
Yes. That is a mistake. I am removing it. – Shahab Apr 15 '13 at 14:25
So would it be 12!/4!4!4!? – hjg hjg Apr 15 '13 at 14:32… – Shahab Apr 15 '13 at 14:34
(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

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