Question

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.
$\begin{array}{cccc}& Column-I& & Column-II\\ \left(A\right)& Thenumberofwaysinwhichallthese& \left(P\right)& 3\\ & coinscanbedistributedifallcoins& & \\ & areidenticalbutallpotsaredifferent& & \\ \left(B\right)& Thenumberofwaysinwhichall& \left(Q\right)& 23\\ & thesecoinscanbedistrbutedifallcoins& & \\ & aredifferentbutallpotsareidentical& & \\ \left(C\right)& Thenumberofwaysallthesecoinscan& \left(R\right)& 15\\ & bedistributedsuchthatnopotisemptyifall& & \\ & coinsaredifferentbutallpotsareidentical& & \\ \left(D\right)& Thenumberofwaysallthesecoinscanbe& \left(S\right)& 12\\ & distributedsuchthatnopotisemptyifallcoins& & \\ & areidenticalbutallpotsaredifferent& & \end{array}$

• A. $A\to R,B\to q,C\to s,D\to P$
• B. $A\to q,B\to r,C\to s,D\to P$
• C. $A\to q,B\to s,C\to r,D\to P$
• D. $A\to p,B\to s,C\to r,D\to q$

Answer 1

The correct option is A $A\to R,B\to q,C\to s,D\to P$

(A) Required number of ways= ${}^6{C}_2$
(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 =$4c_4=1$
For the second case No of selections = $4c_3\times 1c_1=4$
For the third case No of selections =$4c_2\times 2c_2=6$
For the fourth case No of selections =$4c_2\times 2c_1=12$
(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 $4c_2$ ways and rest can be put in $\frac{2c_1\times 1c_1}{2!}$
Required number of distributions = $4c_2\times 2c_1=12$
(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 $3c_1$ ways.Hence, the required number of distributions = $3_{c_1}\times 1=3$