How many ways are there to pick a selection of $\$1$ worth of identical pennies, $\$1$ worth of identical nickels, $\$1$ worth of identical dimes if you select a total of $16$ coins?
Soln: At first the quantities threw me off when they provided too much information, but the key word is identical which means I do not have to choose between the elements. That being said I applied the bars and stars method arriving at: $C(16 + 3 – 1, 16) =
C( 16 + 3 – 1, 2)$
but looking at their solution they had: $C(16 + 3 – 1, 16) – C(5 + 3 – 1, 5)$
Is it a possible solution typo? I don't see where or why they subtracted, unless they wanted to ask "how many of a certain amount of coin given that there must be at least of coin X"
Best Answer
Summarising my comment and lulu's various comments:
If you could use any number of pennies, nickels and dimes, then there would, as you say, be $C(16 + 3 - 1, 16) = C( 16 + 3 - 1, 2)=153$ possibilities of choosing $16$ of them, using a stars-and-bars argument.
But you only have ten dimes ($\$1$ worth of identical $10c$ coins). So you need to subtract cases with eleven or more dimes. The number of cases with $11$ or more dimes is $C((16-11) + 3 - 1, (16-11)) = C( 5+ 3 - 1, 5)=21$, again using a stars-and-bars argument.
You do not have an additional constraint from the $20$ nickels or $100$ pennies as you are only selecting $16$ coins.
So the answer is $C(16 + 3 - 1, 16)-C( 5+ 3 - 1, 5) = 132$ possibilities.