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Evaluate 1. (n-2)! x (n-1) 2. n x (n-1)!
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you know factorial of a number?
@JaneReducto reply :)
let's see, 3! = factorial of 3 = 1*2*3 { multiplication of numbers from 1 to given number, here 3} that way 5! = 1*2*3*4*5 now n! = 1*2*3*........(n-2)*(n-1)*n rigth?
we generally have these below right 4! = 1*2*3*4 5! = 1*2*3*4*5 5! = (1*2*3*4)*5 = 4!*5
for your second problem, if you take n = 5, thne n-1 will be 4 so 4!*5 is actually 5 right? so (n-1)!*n will be n!
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Please note that, using the definition of factorial, we have: \[(n-2)!*(n-1)=(n-1)!\] since (n-1) is next of (n-2), and similarly,we can write: \[(n-1)!*n=n!\] since n is the next of (n-1)
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