+0

# Hard Counting Problem (IMO)

+1
669
4
+169

An ancient human tribe had a hierarchical system where there existed one chief with 2 supporting chiefs (supporting chief A and supporting chief B), each of whom had 2 equal, inferior officers. If the tribe at one point had 10 members, what is the number of different ways to choose the leadership of the tribe? That is, in how many ways can we choose a chief, 2 supporting chiefs, and two inferior officers reporting to each supporting chief?

-----

Okay, so I tried doing this problem like $$10\cdot\binom{9}{2}\cdot \binom{7}{2}\cdot\binom{5}{2} = \boxed{75600}$$, but I realized that doesn't work because the 2 supporting chiefs are distinct, meaning you can't just do 9 choose 2 like that. In this problem, the 10 is the same, but I don't really know how to go from there? Any hints?

Aug 8, 2020

#1
+1082
+5

Do you really need to use "choose"? Can you form a simple premutation?

P.S. I think you are overthinking...

:)

Aug 8, 2020
#2
+169
+1

We have 10 choices for the first chief, no doubt about that

then 9 choices for the supporting chief A

ASSUMING A officers are inferior to b, we have 8C2 ways

Then 6 for support B

then 5C2 for b officer?

Really? that's it?

Aug 8, 2020
#3
+1082
+4

Not exactly, but you are getting closer! Let's think about it case by case.

So, you are right that there is one chief, so that is 10.

Now, don't go ahead and choose the officers first! First, choose your two supporting chiefs. So that is 9C2.

Next, we choose our officers. It is not 5C2, since 1) we do not have 5 left (9-2=?) and 2) you need 4 officers, not 2 officers.

Can you go from here?

:)

ilorty  Aug 8, 2020
#4
+169
+1

I got it! Thanks!

(also choosing what first doesn't matter, i just learned that :o)

Inspirational  Aug 9, 2020