# How many ways to order such that one group comes first?

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 $$0.25*1,320 = 330$$.

• 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. Commented Sep 2, 2021 at 14:21

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

$${n+k-1}\choose{k-1}$$
Where there are $$n = 7$$ yellow cards to be placed into the $$k = 5$$ bins. This gives $$330$$, matching your work.