For a positive integer n, the number n! is defined to be

Question

For a positive integer n, the number n! is defined to be n(n - 1)(n - 2) ... (1). For example, 4! = 4(3)(2)(1). What is the value of 5! - 3!?

(A) 120
(B) 114
(C) 20
(D) 15
(E) 2

Abhishek009 · Answer

5! - 3!

=  5*4*3! - 3!

=  3!( 5*4 - 1)

=  3*2 ( 20 - 1 )

=  6*19

=  114 , Answer must be (B)

ocelot22 · Answer

5!-3! = 5*4*3*2*1-3*2=3*2(5*4-1) = 6*19=114. This probably is a sub 600 level question though, not 600-700.

Jason256 · Answer

For a positive integer n, the notation n! represents the factorial of n. The factorial of a positive integer n is calculated by multiplying all positive integers from 1 to n together.

Mathematically, n! can be defined as:

n! = n × (n - 1) × (n - 2) × ... × 3 × 2 × 1

For example, if n = 5, then 5! (read as "5 factorial") would be calculated as:

5! = 5 × 4 × 3 × 2 × 1 = 120

In general, n! grows very quickly as n increases. The factorial function is commonly used in various mathematical calculations, combinatorics, and probability theory.

unraveled · Answer

Since 3! is common in 5! - 3! taking is out
3! (5*4 - 1) give a number that is multiple of 6(3! = 3*2 = 6)

Thus, C,D and E are out.
As that number has to be less by 6 only 114 satisfies the condition.

Answer B.

ManifestDreamMBA · Answer

This can be solved simply by subtraction, 120-6 = 114, as we know 5!= 120 and 3! = 6. 
Though, even, I followed the same approach as everyone else above first, but was thinking could have just used the values directly.

For a positive integer n, the number n! is defined to be

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For a positive integer n, the number n! is defined to be

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