#2**+10 **

**Badinage**, you are right.

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

#1**+8 **# ðŸŒ¹ ðŸŒ¹ ðŸŒ¹ ðŸŒ¹ ðŸŒ¹ ðŸŒ¹ ðŸŒ¹ ðŸŒ¹

**Yes, that's the answer your teacher is seeking.**

**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.

.

Badinage May 2, 2015

#2**+10 **

Best Answer

**Badinage**, you are right.

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