Question

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

In how many ways all these coins can be distributed if out of 4 coins 2 coins are identical and all pots are different?

• A. 45
• B. 27
• C. 54
• D. None of these

Answer 1

The correct option is C 54

The possible arrangements are as follows:
(4, 0, 0) $\to$ can be done in 3 ways
Let the 4 coins be A, A, B and C then all these 4 coins as a single packet of coins can be arranged in 3 different pots in 3 ways.
(3, 1, 0) $\to$ can be done in 18 ways
out of A, A, B and C, we can select one coin in 3 ways i.e, either A or B or C.
Now we can arrange this selected coin in any 3 pots in 3 ways and the remaining 3 coins as a single packet of coins can be arranged in remaining 2 pots in 2 ways.
Hence, the required number of ways $=3\times 3\times 2=18$
(2, 2, 0) $\to$ can be done in 12 ways
There are two possible cases:
(i) (A, A), (B, C)
(ii) (A, B), (A, C)
In each of two cases we assume that there are two packets of coins which can be arranged in $3\times 2\times 2=12$
(2, 1, 1) $\to$ can be done in 21 wys.
There are 4 possible cases:
(i) (A, A), (B), (C) (ii) (A, B), (A), (C)
(iii) (A, C), (A), (B) (iv) (B, C), (A), (A)
For the first 3 cases in each case, all the 3 packers of coins can be arranged in 3! ways.
Hence, the number of arrangements $=3\times 3!=18$
Now in the fourth (iv) case, two coins are identical so the third packet of coins can be arranged in 3 pots in 3 ways.
Hence, the total number of arrangements = 18 + 3 = 21
Thus the required number of ways = 3 + 18 + 12 + 21 = 54 ways