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11 February, 17:22

How do you simplify (n!) ^2 / (n+1) ! (n-1) !

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  1. 11 February, 17:57
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    We haven n! = (n-1) ! x n and (n+1) ! = n! x (n + 1);

    Then, (n!) ^2 = n! x n! = n! x (n-1) ! x n;

    And (n+1) ! (n-1) ! = n! x (n + 1) x (n-1) !;

    Finally, [n! x (n-1) ! x n] / [n! x (n + 1) x (n-1) !] = (n+1) / n;
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