SOLUTION: Show that r(nCr) = n[(n-1)C(r-1)]

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Question 920291: Show that r(nCr) = n[(n-1)C(r-1)]
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
r%2A%28nCr%29%22%22=%22%22n%2A%28%22%28n-1%29%22%5B%22%22%5D%2AC%28r-1%29%29
By definition ACB+=+A%21%2F%28B%21%28A-B%29%21%29

r%2A%28n%21%2F%28r%21%28n-r%29%21%29%29%22%22=%22%22n%2A%28%28n-1%29%21%2F%28%28r-1%29%21%28n-1-%28r-1%29%29%21%29%29

r%2A%28n%21%2F%28r%21%28n-r%29%21%29%29%22%22=%22%22n%2A%28%28n-1%29%21%2F%28%28r-1%29%21%28n-1-r%2B1%29%21%29%29

r%2An%21%2F%28r%21%28n-r%29%21%29%22%22=%22%22%28n%2A%28n-1%29%21%29%2F%28%28r-1%29%21%28n-r%29%21%29

Write the r! in the denominator on the left as r(r-1)!
Write the n*(n-1)! in the numerator on the right as n!

r%2An%21%2F%28r%2A%28r-1%29%21%28n-r%29%21%29%22%22=%22%22n%21%2F%28%28r-1%29%21%28n-r%29%21%29

Cancel the r's on the left:

cross%28r%29%2An%21%2F%28cross%28r%29%2A%28r-1%29%21%28n-r%29%21%29%22%22=%22%22n%21%2F%28%28r-1%29%21%28n-r%29%21%29

n%21%2F%28%28r-1%29%21%28n-r%29%21%29%22%22=%22%22n%21%2F%28%28r-1%29%21%28n-r%29%21%29

Both sides are now the same.

Edwin