In how many ways can three different awards be distributed among the 17 people in our school, if a.) No student may receive more than one award? B) there is no limit on the number of awards won by one student?   Explain please

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a) You need to use the formula for the number of ways of choosing r items/people from n items/people

`C(n,r) = (n!)/(r!(n-r)!)`

We say 'n choose r'

Here `n=17` and `r=3` so the number of ways is

`C(17,3) = (17!)/(3!14!) = (17.16.15.14.13...3.2.1)/((3.2.1)(14.13.12...3.2.1))=(17.16.15)/(3.2.1)`

Cancelling on the top and bottom gives `C(17,3) = 17.8.5...

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a) You need to use the formula for the number of ways of choosing r items/people from n items/people

`C(n,r) = (n!)/(r!(n-r)!)`

We say 'n choose r'

Here `n=17` and `r=3` so the number of ways is

`C(17,3) = (17!)/(3!14!) = (17.16.15.14.13...3.2.1)/((3.2.1)(14.13.12...3.2.1))=(17.16.15)/(3.2.1)`

Cancelling on the top and bottom gives `C(17,3) = 17.8.5 = 680`  

b) Now we have `C(17,3)` ways of choosing 3 individuals each receiving 1 prize, plus `C(17,2)` ways of choosing 2 individuals receiving 2 prizes and 1 prize respectively, plus `C(17,2)` ways of choosing 2 individuals receiving 1 prize and 2 prizes respectively, plus `C(17,1)` ways of choosing 1 individual receiving all 3 prizes.

So we have the number of ways is

`C(17,3) +2C(17,2) + C(17,1) = 680 + 2((17!)/(2!15!)) + (17!)/(1!16!)`

` = 680 + 2((17.16.15.14.13...3.2.1)/((2.1)(15.14.13...3.2.1))) + (17.16.15.14.13...3.2.1)/((1)(16.15.14.13...3.2.1))`

` ``= 680 + 2((17.16)/(2.1)) + (17)/(1) = 680 + 2.17.8 + 17 = 680 + 272 + 17 = 969`

a) 680 ways

b) 969 ways

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