+0

+1
98
6
+44

There are a hundred competitors at the National Debating Contest, two from each of the 50 US states. In how many ways can five finalists be chosen if no state may have more than one finalist?

Jan 5, 2021

#1
+117576
+1

Any  1  of the 100  can be chosen first

Next.....we  can  choose  any 1  of the other (100 - 2)  =   98   ( we can't choose  the other  one from the  first state)

Next.....we can choose any of the remaining  (100 - 4)  = 96  (we have chosen two  from two states  and we  can't choose the  remaining two from either state)...so...we only have 96 to choose from

Continuing with this pattern :

Then.... one from the remaining  (100 - 6)  = 94

Then... one from the remaining (100 - 8)  = 92

So

100 * 98 * 96 * 94 * 92    =   8,136,038,400   ways  !!!

THIS IS INCORRECT   !!!

Jan 5, 2021
edited by CPhill  Jan 5, 2021
#2
+44
0

Wait, but according to the official national MATHCOUNTS solutions, the answer is 67,800,320 ways, was there something wrong in your solution?

GR0001  Jan 5, 2021
#3
+117576
+1

Don't know where I made an error....maybe someone else can  give you the  correct answer....sorry!!!

CPhill  Jan 5, 2021
#4
+44
0

That's okay, I'll keep working to see how to arrive at 67,800,320. Thanks anyways though!

GR0001  Jan 5, 2021
#5
+44
+1

Found it!

Pick 5 states out of the 50 possible, 50 choose 5 (2118760)

Choose 1 out of 2 competitors out of each state.

25 because there are 5 states and 2 choices for each.

2118760*32=67800320

GR0001  Jan 5, 2021
#6
+44
+1

Found the answer after working on the problem for a bit:

Pick 5 states out of the 50 possible, 50 choose 5 (2118760)

Choose 1 out of 2 competitors out of each state.

25 because there are 5 states and 2 choices for each.

2118760*32=67800320

Jan 5, 2021
edited by GR0001  Jan 5, 2021