Problem Solving

dumb.doofus wrote:I think the answer should be 360. Login to see the pics..

If someone has a quicker way to solve this.. that would be awesome.. please do share..

For now here is the solution:

Let's divide this into two portions:

A] When the 3 C's are placed together

Let's denote the 3 C's as R.. So the elements are AAABBR

Permutation = 6!/(3!2!) = 60 ------------------- (1)

B] When 2 of the C's are placed together

Let's again denote the 2 C's by R

so now the elements are: AAABBCR
Now again there are two cases:
1. When R is placed as the first element or the last element. In both cases permutation will be same.

So permutation for R _ _ _ _ _ _ _ or _ _ _ _ _ _ R will be



P = (5*5*4*3*2*1)/(3!2!) = 50

So total permutation for first and last position = 50 + 50 = 100 ----- (2)

2. When R is placed anywhere in the middle i.e. 2nd, 3rd, 4th, 5th or 6th position.. permutation will be same..



so P = (5*4*4*3*2*1)/(3!2!) = 40

total permutation = 5 * 40 = 200 --------------------- (3)

Total permutation = (1) + (2) + (3)
= 60 + 100 + 200
= 360

agoyal2 wrote:PAB2706, I am also getting 350 :roll:

Total Arrangements = 8! / (3!*2!*3!) = 560

When 2 C's are considered 1 = 7! / (3!*2!*2!) = 210
(The above also includes when all 3 C's are together...)

So total cases = 560-210 = 350 ??