Answer:
a is the only true one if you meant 6 times 5!
Step-by-step explanation:
Before we being
n!=n*(n-1)*(n-2)*...(3)(2)(1)
Example: 5!=5(4)(3)(2)(1) and 10!=10(9)(8)(7)(6)(5)(4)(3)(2)(1)
Yes that operation is multiplication.
a) Does 6!=6*5!
Let's see
6!=6*5*4*3*2*1
5!=5*4*3*2*1
So 6*5!=6(5*4*3*2*1)=6*5*4*3*2*1=6!
So true!
b) Does 4!+2!=6! ?
4!=4(3)(2)(1)
2!=2(1)
6!=6(5)(4)(3)(2)(1)
Does
4(3)(2)(1)+2(1)=6(5)(4)(3)(2)(1)
24 +2 =720
26=720 (this is not true)
So 4!+2! is not 6!
c) Does 6!/3!=2! ?
6!=6(5)(4)(3)(2)(1)
3!=3(2)(1)
If you divide 6! by 3!, then the factors 3 and 2 and 1 cancel and you are left with 6(5)(4).
So the question becomes is 6(5)(4)=2!
2!=2(1)=2
6(5)(4)=2?
No way! That is saying 120=2 which is not true.
