# Finite Math

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In a student writing contest with 25 entries, 3 essays are selected for first, second, and third place awards and 5 are selected for honorable mention. How many ways can this be done? Thanks.

Ok Gen, for the first place there are 25 choices, for second place 24 choices and for third place 23. There are thus 25-times-24-times-23 permutations for the first three places
Would you agree and do you see why?
That leaves 22 remaining from which we can select 5 in any order. That means there are 22 choose 5 combinations for honorable mention.
The total ways this can be done is the product of these two numbers:
(permutations of 25 with 3) times (combinations of 22 choose 5).
Do you see why?
In general, when order matters think permutations. When it doesn't, think combinations.

Thank you Roger. I do understand the order of the first 3, and the non order of the 5. I just did not know that permutations and combinations could be combined. When I do this I get 363,409,200. This is a huge number.Is this correct?

That looks correct by my calculations.
Both permutations and combinations use factorials and these grow very quickly. So numbers this size are common.

Thank you very much Roger.

Find an equation of the line containing the given pair of points (3,5)and (9,8)

assume that there are 4 dimes, 3 nickels, and 2 quarters.

In how many possible ways can the selection be made so that the value of the coins is at least 25 cents?