Question

If the $(n+1)$ numbers $a,b,c,d,........$be all different and each of them a prime number, then the number of different factors (other than $1$) of $a^m.b.c.d....$ is

• A. $\left(m+1\right)2^n$
• B. $\left(m+1\right)2^n-1$
• C. $\left(m+1\right)2^n+1$
• D. None of these

Answer 1

Answer: (B), (C)

$\left(m+1\right)2^n+1$

Explanation for correct option:

Number of element in $b.c.d...=n$

Choose $a^k$ where $k\in 0,1,2,...,m$ at a time $=\left(m+1\right)$

For every $k,$ number of possible factors from $n$ elements $={}^{n}C_0+{}^{n}C_1+{}^{n}C_2+...+{}^{n}C_n=2^n$

Total factors $=$ Number of possible powers of $a$$\times$every $k$ number of possible factors from $n$ elements

$=\left(m+1\right)2^n$

If factor $1$ is excludes then $=\left(m+1\right)2^n-1$

Hence, option B is correct