Question

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 3 different pots in 3 ways.
(3, 1, 0) $\to$ can be done in 18 ways
out of A, A, B and C we van select one coin in 3 ways i.e, either A or N 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 $3\times 2=12$
(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 cases all the 3 packers of coins can be arranged in 3! ways.
Hence, the number of arrangements $=\times 3!=18$
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