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By Cooljay687 (Cooljay687) on Saturday, May 17, 2003 - 08:40 pm: Edit |

Came from PR SAT

The Tyler Jackson Dance COmpany plans to perform a piece that requires 2 dancers. If there are 7 dancers in the company, how many different pairs of dancers could perform the piece?

I thought it would be just 7 x 2 = 14

Answer turns out to be 21.

When writing out all the possibilities, I do come up with 21. However, is there an arithmetic way to do this? I am stuck!

Thanks.

By Jwood (Jwood) on Saturday, May 17, 2003 - 08:57 pm: Edit |

6+5+4+3+2+1=21

First dancer has 6 possible partners

Second dancer has 5 possible (different) partners

Third dancer has 4 possible (different) partners

etc...

By Sixerztres (Sixerztres) on Saturday, May 17, 2003 - 09:36 pm: Edit |

this is called a combination problem. on your calculator there's a button that says nCr. punch in 7, nCr, 2 and you will get 21. mathematically you do n!/((n-r)!r!) = 7!/(5!2!) = 21.

By Student108 (Student108) on Saturday, May 17, 2003 - 09:46 pm: Edit |

all u need to do on ur calculator is punch in 7 the nCr and then 2 and theres your answer well after you press equal COMBINATIONS of course.....thats why that buttons there...Its prett simple. Whenever they talk about as many combinations..use that

By Ndhawk (Ndhawk) on Saturday, May 17, 2003 - 10:50 pm: Edit |

My method is: 7 x 6/ 2 x 1

By Ndhawk (Ndhawk) on Saturday, May 17, 2003 - 10:50 pm: Edit |

that ends up being sixer's formula but simplified, i just start off doing it the simplified way

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