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  1. How many different collections of six coins can be formed, if each coin may be a cent, a nickel, a dime, a quarter, a half dollar, or a silver dollar?

$\binom{6+6-1}{6} = \binom{11}{6}$

  1. Poker chips come in three colors, red, white, and blue. How many different combinations of ten poker chips are there?

$\binom{10+3-1}{10}$ = $\binom{12}{10}$

I have a question regarding the 2 questions/answers above. From my understanding, the formula: $\binom{n+r-1}{n}$ is used when including 0(not choosing) into the count. However, there is no mentioning including 0 as a choice in the question, hence, why the answer?

The questions/answers are from "math of choice"

Kindly advise

Update:
I've omitted "the other way" where the denominator is n-r to avoid confusing of my question.

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  • $\begingroup$ @RaymondManzoni, thank you for the reply. Maybe, i didn't ask the question right (which I've corrected). I'm asking why use $\binom{n+r-1}{n}$ instead of $\binom{n-1}{r-1}$ since there is no mentioning of including 0 as a choice. $\endgroup$ May 21, 2022 at 7:05
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    $\begingroup$ In the questions, unless it is specifically mentioned that there must be at least one of each coin/color, we take that some may be $0$ $\endgroup$ May 21, 2022 at 7:14

1 Answer 1

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"...if each coin may be a cent, a nickel, a dime, a quarter, a half dollar, or a silver dollar."

There is a (widely accepted) convention, that if the intent of the problem composer is that at least one of each type of coin be selected, then the problem composer will include the constraint

"At least one of each type of coin will be chosen.

For the present present problem, there is the added factor that the assumption of this unstated premise would make the problem close to pointless, because there would then be only $1$ solution possible. That is, you would then have to have exactly one of each type of coin.


"...How many different combinations of ten poker chips are there?"

Again, it is generally accepted that if the problem composer intended that at least one of each type of chip must be present, he would include a constraint like:

"One of each type of poker chip must be present."

In rebuttal...

However, unlike the first problem, this problem would still be meaningful, if the unstated additional premise was assumed. Also, it is (at least somewhat) unclear just how widespread the convention is that if the additional premise is not explicitly stated, that it is acceptable (for example) to have all $10$ poker chips be the same color.

There is also an additional complication. Sometimes, reality intrudes on the convention. For example, suppose that you are going to be distributing $10$ indistinguishable gifts to $3$ children. In how many different ways can this be done?

Here, if it is not explicitly stated that each child must get at least one gift, then I (for one) would explicitly ask whether each child must receive at least one gift.

So, in this alternative problem, reality has intruded on the generally accepted convention.

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