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I am completing a practice math test, so I know the answer to the following problem (10), but I'm not sure how you would get it, and how it would apply for a different situation.

A group of five friends has two tickets to the ball game. How many different combinations of these five friends can use the tickets?

I have experimented and come to the answer using $4+3+2+1$, but I am unsure if that's anywhere close to being right, and how it would apply if there were a different number of tickets, etc.

Thanks for any help

Sri-Amirthan Theivendran
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DragonzInsanity
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  • It is correct, however there is a much faster much easier way of seeing why it would be $10$ instead of adding several numbers together by looking at it using binomial coefficients. You have five people. You want to choose two of them to get tickets. The number of ways this can be done then is "five choose two", i.e. $\binom{5}{2}=\frac{5!}{2!(5-2)!}=10$ ways. – JMoravitz Oct 13 '17 at 21:32
  • @JMoravitz Thanks for the help – DragonzInsanity Oct 13 '17 at 21:35
  • Just to make sure that you followed my previous comment and understood, can you now answer the question "A group of eight friends have three tickets to the ball game. How many different combinations of these eight friends can use the tickets?" – JMoravitz Oct 13 '17 at 21:46
  • @JMoravitz The answer would be 56 ways. Thanks again. – DragonzInsanity Oct 14 '17 at 23:24

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