# Probability of C+A+R+D=35

• Aug 5th 2012, 10:19 PM
tykim
Probability of C+A+R+D=35
I was asked to solve the below. Could I get help?

Four letters of C, A, R and D have its own natural number: 1,2,3,4,5,.... Sum of the four letters, C+A+R+D=35.

A. Total number of possible pairs?
B. Probability of C=32 or A=32 or R=32 or D=32 ?
• Aug 5th 2012, 10:52 PM
richard1234
Re: Probability of C+A+R+D=35
A: Do you know the "balls and boxes" or "divider" technique?

Basically, we subtract 1 from each number so that C,A,R,D are nonnegative integers with C+A+R+D = 31. Now we have 31 units and three "dividers" to split with. Any permutation of the 31 units and three dividers results in four numbers C,A,R,D such that C+A+R+D = 31. The total number of ways is $\binom{34}{3} = \frac{34!}{31!3!}$.

B: Let C = 32, find the number of ways such that A+R+D = 2. Multiply by 4 to account for A = 32, etc.
• Aug 5th 2012, 11:07 PM
tykim
Re: Probability of C+A+R+D=35
Thanks richard1234 for the quick reply.

TYKim
• Aug 6th 2012, 04:59 PM
tykim
Re: Probability of C+A+R+D=35
Dear Richard1234,

I can not get it with problem B. Could you give me further information?

Thanks.
• Aug 6th 2012, 05:48 PM
richard1234
Re: Probability of C+A+R+D=35
Okay. Suppose C = 32. How many possible ways are there such that A+R+D = 3 (sorry, not 2)? This one's pretty easy to brute force.

Then suppose A = 32. You want the number of ways such that C+R+D = 3. However this is the same as with the C = 32 case. That's why you multiply by 4.