# Combination problem with identical objects

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

• 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.) – saulspatz Jan 30 at 19:44

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 $$54$$.