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 ?
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 ?
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 $\displaystyle \binom{34}{3} = \frac{34!}{31!3!}$.
B: Let C = 32, find the number of ways such that A+R+D = 2. Multiply by 4 to account for A = 32, etc.
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.