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# If three runners receive medals how many ways can the three medalists be chosen out of a field of 28?

0
584
3
+8

Is it 28 x 27 x 26?

May 2, 2015

#2
+95179
+10

There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd   That is 28*27*26

BUT it doesn't matter what order they are chosen.  so you have to divide by 3! = 6

$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$

this is how I would normally do it

28C3

$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$

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May 2, 2015

#1
+520
+8

However,  as a challenge extension you could extend this to include ties, e.g., from 2 up to 28 ties for first, up to 27 ties for second, etc.     I'm not sure how to do this, though, and it sounds complicated.

# ðŸŒ¹  ðŸŒ¹  ðŸŒ¹  ðŸŒ¹  ðŸŒ¹  ðŸŒ¹  ðŸŒ¹  ðŸŒ¹

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May 2, 2015
#2
+95179
+10

There are 28 ways for the first, 27 ways for the second and 26 ways for the 3rd   That is 28*27*26

BUT it doesn't matter what order they are chosen.  so you have to divide by 3! = 6

$${\frac{{\mathtt{28}}{\mathtt{\,\times\,}}{\mathtt{27}}{\mathtt{\,\times\,}}{\mathtt{26}}}{{\mathtt{6}}}} = {\mathtt{3\,276}}$$

this is how I would normally do it

28C3

$${\left({\frac{{\mathtt{28}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{28}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{3\,276}}$$

Melody May 2, 2015
#3
+520
+5

Yes, you are right, Melody. The question doesn't say 1st, 2nd and 3rd places. Maybe there are 3 runners who are to be awarded a medal for being a comically costumed competitor for St Patrick's Day, or some such.

May 3, 2015