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I came across a question that asked: Two punts can each hold 6 people. A party of 10 wishes to use these punts. In how many different ways can the party be divided? Assume that each member of the party is distinct.

Please could I check how to go about calculating the number of combinations?

Would I be correct if I say the answer is the following :

$^{10}C_6 + ^{10}C_5 + ^{10}C_4 = 672$

Key Flex
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Ari
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    Are the punts considered distinct? If yes, then pick which size 4,5, or 6 subset of the people use the left punt. The remaining people use the right. If the punts are not considered distinct, then take the youngest person. Decide what 3,4, or 5 other people go to the same punt as them. The remaining people go to the other. – JMoravitz Jul 20 '18 at 23:50
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    @JMoravitz if they were all born the same day, take the one with the bigger shoe size. – Arnaud Mortier Jul 20 '18 at 23:57
  • The example of taking the youngest is just an example. The point of course being that given they are all distinct and given that they are only countably many of them, there must exist some way of ordering them. Pick any such order and treat the "smallest" according to that order as special. Ordering them according to age is just one natural example that occurs frequently in real life. – JMoravitz Jul 20 '18 at 23:59
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    "Would I be correct if I say the answer is 10C6 + ... = 672?" Yes you would, assuming you make the assumption that the punts are considered distinct (e.g. a west punt and an east punt) and it is considered relevant which field people play on. – JMoravitz Jul 21 '18 at 00:01
  • Many thanks, @JMoravitz – Ari Jul 21 '18 at 09:28
  • @JMoravitz In my experience, a punt is a kind of boat. Of course that has no effect on the answer to the question. – David K Jul 21 '18 at 10:30
  • @Davidk ah, I was guessing it was like how fields are referred to as pitches in some sports by some countries. – JMoravitz Jul 21 '18 at 15:27

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