Question

There are 3 pots and 4 coins. All these coins are to be distributed into these pots where any pot can contain any number of coins.

In how many ways all these coins can be distributed such that no pot is empty if all coins are different but all pots are identical?

• A. 16
• B. 6
• C. 42
• D. 21

Answer 1

The correct option is B 6

Since no box is empty and all pots are identical so the possible case is (1, 1, 2).
But since all the coins are different, then the 2 balls can be selected in ${}^4{C}_2$ ways and rest can be put in $\frac{{}^2{C}_1{\times }^1{C}_1}{2!}$ ways.
Hence, the required number of distribution
$=\frac{{}^4{C}_2{\times }^2{C}_1{\times }^1{C}_1}{2!}=6$