Find the total number of ways in which a beggar can be given at least 1 dollar from four 25 cent coins, three 50 cent coins and two 1 dollar coins

My Attempt

Total: 9

N-[ N(nothing) + N(one 25 cents) + N(two 25cents) + N(three 25 cents) + N(one 50 cents) + N(one 50 cents + one 25 cents)]=

$$ 2^9-[1+ {}^4C_1+{}^4C_2+{}^4C_3+{}^3C_1+{}^3C_1.{}^4C_1 ]=512-[1+4+6+4+3+12] $$ Combination problems with identical objects are always seem to be a headache for me i think, What am I thinking wrong here ?

And what is the easiest way to approach the problems like this ?

The solution given in my reference is $54$ ways

  • $\begingroup$ First of all, you've missed the possibility of a quarter and a fifty-cent piece. Secondly, you're treating all the coins as distinct, which I doubt the problem intends. (That is, you count $4$ ways he can get $3$ quarters, but the problem views them as the same.) $\endgroup$ – saulspatz Jan 30 at 19:44

As I noted in a comment, the problem is that you are looking at individual coins, whereas the problem views coins of the same denomination as indistinguishable.

We can give the beggar from $0$ to $4$ quarters, so there are $5$ possibilities. Similarly, there are $4$ possibilities for the $50$-cent pieces and $3$ for the dollar coins, so $5\cdot4\cdot3=60$ possibilities in all.

Subtracting the $6$ possibilities you found for combinations that don't come to at least a dollar, we get $54$.

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