Question

In how many ways can $6$ prizes be distributed equally among $3$ persons?

• A. ${}^6{C}_2{\times }^4{C}_2$
• B. ${}^6{P}_2{\times }^4{P}_2$
• C. $3$
• D. $3^6$

Answer 1

The correct option is A ${}^6{C}_2{\times }^4{C}_2$

There are $6$ prizes have to be evenly distributed, i.e. $2$ prizes person.

For the first person we have to select $2$ from $6$,

$\therefore$ No. of combination$={6C}_2.$

$4$ prizes are left for second person$={4C}_2$ & for third person $={2C}_2$.

$\therefore$ Total required no. of ways$={6C}_2\times {4C}_2\times {2C}_2={6C}_2\times {4C}_2\times 1={6C}_2\times {4C}_2.$