Question

If $\angle 15=2^{P_1}{3}^{P_2}{5}^{P_3}{7}^{P_4}{11}^{P_5}{13}^{P_6}then{\sum }_{r=1}^6{P}_r$ is

• A. 24
• B. 23
• C. 22
• D. 21

Answer 1

The correct option is A 24

$15!=2^{11}\times 3^6\times 5^3\times 7^2\times {11}^1\times {13}^1$ $\therefore {\sum }_{r=1}^6{P}_r=11+6+3+2+1+1=24$