write each expression without using the factorial symbol.
1)(n-3)!
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n!
2) (n+4)!
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(n+2)!
I will do one. Take the greater number, and expand it. for instance, the first
(n-3)!/(n*(n-1)(n-2)(n-3)!)
the (n-3)! terms divide out..
1/(n)(n-1)(n-2)
then the second one would be
(n+4)!/((n+1)(n+2)(n+3)(n+4)(n+2)!)
=
1/(n+1)(n+2)^2 (n+3) ?
To write each expression without using the factorial symbol, we need to understand that n! (read as n factorial) represents the product of all positive integers from 1 to n.
1) To write (n-3)!/n! without using the factorial symbol, we can expand the numerator and denominator.
(n-3)! = (n-3) * (n-4) * (n-5) * ... * 3 * 2 * 1
n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1
By dividing the two expressions, most of the terms cancel out, leaving:
(n-3)!/n! = [(n-3) * (n-4) * ... * 3 * 2 * 1] / [n * (n-1) * (n-2) * ... * 3 * 2 * 1]
All the terms from (n-2) down to 2 and 1 cancel out:
(n-3)!/n! = (n-3) / [n * (n-1)]
2) To write (n+4)!/(n+2)! without using the factorial symbol, we can expand the numerator and denominator.
(n+4)! = (n+4) * (n+3) * (n+2) * (n+1) * n * ... * 3 * 2 * 1
(n+2)! = (n+2) * (n+1) * n * ... * 3 * 2 * 1
By dividing the two expressions, most of the terms cancel out, leaving:
(n+4)!/(n+2)! = [(n+4) * (n+3)] / [(n+2) * (n+1) * n * ... * 3 * 2 * 1]
All the terms from n down to 3, 2, and 1 cancel out:
(n+4)!/(n+2)! = (n+4) * (n+3)