These are three pots and four coins. All these coins are to be distributed into these pots where any pot can contain any number of coins.
Column1Column2(A)The number of ways in which all these coins can be(P)3distributed if all coins are identical but all pots are different(B)The number of ways in which all these coins can be(Q)6distributed if all coins are different but all pots are identical(C)The number of ways all these coins can be distributed such(R)15that no pot is empty if all coins are different but all pots areidentical.(D)The number of ways all these coins can be distributed(S)14such that no pot is empty is all coins are identical but all pots are different.
AR, BS, CQ, DP
A → R,B → S,C → Q,D → P.
(A)Required number of ways= 6C2
(B)Since pots are identical,there will be 4 cases (4,0,0),(3,1,0),(2,2,0),(2,1,1) but all the coins are different hence selection of coins matters.
For the first case no. of selections =4C4=1
For the second case No.of selection=4C3×1C1=4
For the third case No.of selection=4C2×2C22!=3
For the third case No.of selection=4C2×2C1×1C12!=6
(C) 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, the 2 balls can be selected in 4C2 ways and rest can be put in 2C1×1C12!
Required number of distributions = 4C2×2C1×1C12!=6
(D) Since no pot is empty and all coins are identical the possible case is (1,1,2). But since all three pots are different hence. a pot (which contains 2 coins together) can be selected in 3C1 ways. Hence,the required number of distributions=3C1×1=3