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This question arose while playing a board game that uses cards. Imagine we have 11 cards total. 3 are red (R), 1 is blue (B), and 7 are yellow (Y). How many ways can we randomly order the 11 cards so that B comes before all 3 R?

Intuitively, I know the 1 B card will come before all 3 R cards 25% of the time, and this helps me check my answer. I thought my method was correct, but my count is clearly too high because it gives a ~27% probability.

I'm interested in a concise way to solve this. I've posted my work below to show an honest attempt, but it has two shortcomings: it's wrong, and it seems overly complicated. After reading this answer I think there might be a way to consolidate my terms using a summation (after all, I am partitioning the event space $A$), but I couldn't figure out how to do that.



My solution:

The length of the sample space is $$|S|=\frac{11!}{(3!)(7!)}=1,320$$ because there are 11 cards total, but the 3 reds aren't distinct, and the 7 yellows aren't distinct.

I conceptualized the event $A$ as the subspace of $S$ where the blue/red cards are ordered B R R R (if we temporarily ignore the yellows). There's only 1 distinct way to produce that arrangement. So, I tried to find $|A|$ by figuring out how many ways to arrange the 7 Y into _B_R_R_R_ where each of the 7 yellows can go into any of the underscores _.

We can take the 7 yellows and split them into a single group containing 7. We could put that single group (7Y) into 5 spots (the first _, the second _, etc.). Or we could break our 7Y into 2 groups (6Y & 1Y). The number of distinct, ordered ways to arrange those 2 groups in our 5 _ spots is $P_{5, 2}$. Or we could break our 7Y into three groups (5Y, 1Y, 1Y). Those 3 groups can be arranged distinctly across the 5 _ spots in $P_{5,3}$ $/$ $2!$ ways.

I did this for all possible disjoint groupings and got the results below. For each grouping, I've counted the number of distinct ways those groups can be placed in the 5 _ spots.

  • break 7Y into 1 group:
    • 7Y ($P_{5, 1}$)
  • break 7Y into 2 groups:
    • 6Y, 1Y ($P_{5, 2}$)
    • 5Y, 2Y ($P_{5, 2}$)
    • 4Y, 3Y ($P_{5, 2}$)
  • break 7Y into 3 groups
    • 5Y, 1Y, 1Y ($P_{5, 3}$ $/$ $2!$)
    • 4Y, 2Y, 1Y ($P_{5, 3}$)
    • 3Y, 3Y, 1Y ($P_{5, 3}$)
    • 3Y, 2Y, 2Y ($P_{5, 3}$ $/$ $2!$)
  • break 7Y into 4 groups
    • 4Y, 1Y, 1Y, 1Y ($P_{5, 4}$ $/$ $3!$)
    • 3Y, 2Y, 1Y, 1Y ($P_{5, 4}$ $/$ $2!$)
    • 2Y, 2Y, 2Y, 1Y ($P_{5, 4}$ $/$ $3!$)
  • break 7Y into 5 groups
    • 3Y, 1Y, 1Y, 1Y, 1Y ($P_{5, 5}$ $/$ $4!$)
    • 2Y, 2Y, 1Y, 1Y, 1Y ($P_{5, 5}$ $/$ $2!3!$)

When I sum all of the parentheses, I get $$P_{5,1} + 3P_{5,2} + 2P_{5,3} + \frac{2P_{5,3}}{2!} + \frac{2P_{5,4}}{3!} + \frac{P_{5,4}}{2!} + \frac{P_{5,5}}{4!} + \frac{P_{5,5}}{2!3!} = 360$$

But I'm over-counting by 30. The correct answer should be $0.25*1,320 = 330$.

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  • $\begingroup$ It just occurred to me that there might be an easier way to count if we treat the 3 R's as distinct (R1, R2, R3), and treat the 7 Y's as distinct (Y1, Y2...). I'm probably using 11!/(3!7!) because that's the method I just learned in my stats class. If you see a simpler solution that treats the problem differently ("how many distinct arrangements are there if the R's and Y's are distinct"), that'd be great too, and I should be able to adapt it to the way I've thought about it. $\endgroup$
    – jdcrossval
    Sep 2, 2021 at 14:21

1 Answer 1

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The error in your solution is that you are missing a factor of $1/2!$ in your term for (3Y, 3Y, 1Y).

This is generally a type of "stars and bars" combinatorics question. As you considered, we essentially have 5 "bins" formed by the 4 B and R cards, and want to put the 7 Y cards into these spaces.

__B__R__R__R__

By arguments you can follow in the linked page, we arrive at their Theorem 2. The total number of possible arrangements is

$$ {n+k-1}\choose{k-1} $$

Where there are $n = 7$ yellow cards to be placed into the $k = 5$ bins. This gives $330$, matching your work.

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