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Four friends have an arrangement that on each of $20$ days, one of them will pay for lunch for everyone. If one of the friends volunteers to pay for lunch on $10$ of these days, how many ways are there for the friends to buy these lunches? (Assume the days are distinguishable.)

could you please help me with this question? This is what I got but I am not sure if it is correct: $$^{20}C_{10}* ^9C_2$$

where the first term counts the number of ways to choose $10$ specific days out of $20$ distinguishable days, and the second term counts the number of ways to distribute the remaining $10$ days among the other $3$ friends using the stars and bars technique.

TShiong
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Sara
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    Here's a MathJax tutorial :) – Shaun Apr 18 '23 at 20:20
  • OP, you haven't made apparent the fact whether the friend that volunteers to pay for $10$ of these days ONLY pays for these $10$ days and on the remaining days he will never pay, or he pays for $10$ of these days and on the remaining days he may or may not consider paying. – anon Apr 21 '23 at 07:59

1 Answers1

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The $~\displaystyle \binom{20}{10}~$ factor is good.

So, reserve this factor, and now assume, without loss of generality, that Person-1 pays for Dinners 1 thru 10 (only).

So, the problem has reduced to determining how many distinct ways that there are that the (distinguishable) persons 2 thru 4 can pay for the distinguishable days 11 thru 20.

Stars and Bars is wrong here, specifically because days 11 thru 20 are distinguishable.

That is, consider the specific solution:

$x_2 = 4, ~x_3 = 3, x_4 = 3 \implies x_2 + x_3 + x_4 = 10.$

Here, $~x_i~$ corresponds to the number of days paid for by Person-i.

However, if Person-2 is paying for $~4~$ days,
you have to distinguish between
$~\color{red}{\text{which four days that Person-2 is paying for}}.$

Since there has been no requirement that any of Persons 2 thru 4 have to pay for at least 1 day, you have $~3~$ choices for each day, as to who pays.

Therefore the second factor is $~(3)^{10}.$

So, the overall computation is

$$\binom{20}{10} \times (3)^{10}.$$

Note

If you add the complication that each of persons 2 thru 4 must pay for at least 1 dinner, then you can complete the problem via Inclusion-Exclusion.

See this article for an introduction to Inclusion-Exclusion. Then, see this answer for an explanation of and justification for the Inclusion-Exclusion formula.

For any set $~E~$ with a finite number of elements, let $~|E|~$ denote the number of elements in the set $~E.~$

Let the set $~S~$ denote the collection of all possible ways that persons 2 thru 4 can pay for days 11 thru 20, without any regard for whether any person pays for at least 1 day.

For $~i \in \{2,3,4\},~$ let $~S_i~$ denote the subset of $~S~$ where the constraint that person-i must pay for at least 1 day is violated.

Then, the second factor, which will eventually be coupled with the $~\displaystyle \binom{20}{10}~$ factor, is represented by

$$|S| - |S_1 \cup S_2 \cup S_3|.$$

By symmetry, and from Inclusion-Exclusion theory, this equals

$$|S| - 3|S_1| + 3|S_1 \cap S_2| - |S_1 \cap S_2 \cap S_3|$$

$$= 3^{10} - \left[ ~3 \times 2^{10} ~\right] + \left[ ~3 \times 1^{10} ~\right] - \left[ ~1 \times 0^{10} ~\right].$$

user2661923
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