Let n be defined as n-2-n-1- what is the value of 7-3 -

The correct option is B 8.4Given, box n is equal to (n+2)!(n 1)!=(n+2)(n+1)n We get box 7 is equal to 9 × 8 × 7 =504 We get box 3 is ...

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Standard VII

Mathematics

Fraction as an Operator 'Of'

Let n be de...

Question

Let $n$ be defined as $\frac{(n + 2)!}{(n - 1)!}$ , what is the value of $\frac{7}{3}$ ?

4.4

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Solution

The correct option is B 8.4
Given, box $n$ is equal to $\frac{(n + 2)!}{(n - 1)!} = (n + 2) (n + 1) n$
We get box $7$ is equal to $9 \times 8 \times 7 = 504$
We get box $3$ is equal to $5 \times 4 \times 3 = 60$
Therefore, If we divide those two, we get $\frac{504}{60} = 8.4$ .

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Let n be defined as (n+2)!(n−1)!, what is the value of 73 ?

The correct option is B 8.4Given, box n is equal to (n+2)!(n−1)!=(n+2)(n+1)nWe get box 7 is equal to 9×8×7=504We get box 3 is equal to 5×4×3=60Therefore, If we divide those two, we get 50460=8.4.

Let $n$ be defined as $\frac{(n + 2)!}{(n - 1)!}$ , what is the value of $\frac{7}{3}$ ?

The correct option is B 8.4
Given, box $n$ is equal to $\frac{(n + 2)!}{(n - 1)!} = (n + 2) (n + 1) n$
We get box $7$ is equal to $9 \times 8 \times 7 = 504$
We get box $3$ is equal to $5 \times 4 \times 3 = 60$
Therefore, If we divide those two, we get $\frac{504}{60} = 8.4$ .