The correct option is C 2
Let's first find the value of 2!, 2!=2×1=2
Now, let's simplify and find the values of all the given options one-by-one:
∙ 5!−3!
5!−3!=5×4×3×2×1−3×2×1
Taking out the common factors 3×2×1
⇒5!−3!=(5×4−1)×(3×2×1)
⇒5!−3!=(20−1)×6
⇒5!−3!=19×6
⇒5!−3!=114
As we can see that,
2≠114⇒2!≠5!−3!
Therefore, the value of 5!−3! is not equal to the value of 2!
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∙ (5−3)!
(5−3)!=(2)!=2×1=2
As we can see that,
2=2
⇒2!=(5−3)!
Therefore, the value of (5−3)! is equal to the value of 2!
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∙ 2
By definition only, 2! is equal to 2.
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∙ 21
The value of 2! is 2, hence, it can not be equal to 21.