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I'm being asked 2 questions:

1) In a restaurant, there is exactly one table with $8$ seats unoccupied. How many possibilities are there for $6$ people?

2) In a class with $23$ people, $3$ are called to the teacher. How many possibilities are there?

Now, Question 1 can be solved using: $8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3$

Question 2 however, can only be solved using the binomial coefficient of $23$ and $3$

So my question is: why can't I use the approach used in the first question, with the second one?

3 Answers 3

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1)$\binom{8}{2}\dot{}6!=28\dot{}6!$

2)$\binom{23}{3}=23!/(3!\dot{}20!)=7\dot{}22\dot{}23$

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    Could you elaborate these answers a bit? :) – 2017-01-30
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    1) You can chose any two seats so you have a sub set of size $2$ to the $8$ seats. For each choice you can permute the 6 people in any way, so $6!$. 2) you just need a subset of size $3$ from the $23$ students. There is a mistake: $7\dot{}11\dot{}23$ – 2017-01-30
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    Hmm, wow that is quite an approach for these two questions, never would have thought about going about them like this, thanks for sharing! :) – 2017-01-30
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    @JulioMaldonadoHenríquez Your comment was arguably better than the answer! Put it in there! – 2017-01-31
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You can think in the same way just be careful with the order. In the second question it doesn't matter.

For the first student you have $23$ possibilities, for the second $22$ and for the third $21$. Once the order doesn't matter then the result is:

$$\frac{23\cdot22\cdot21}{3!}$$

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    Thank you! :) That does make a lot of sense! – 2017-01-30
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    You are very welcome! – 2017-01-30
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Because in the second question, if you pick three people $a,b,c$ that is the same as picking three people $c,a,b$ and so forth. In the first question, order matters. So the same approach does not apply to both questions.

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    Thank you, I understand now! :) I'll accept this answer as it was the first(in 10 minutes(not allowed earlier)), although both answers are answering my question. – 2017-01-30