Bunuel wrote:
tania wrote:
In how many different ways can the letters A,A,B,B,B,C,D,E be arranged if the letter C must be to the right of the letter D?
A.1680
B.2160
C.2520
D.3240
E.3360
Can someone explain how I should approach to solve the above problem?
A.1680
B.2160
C.2520
D.3240
E.3360
Can someone explain how I should approach to solve the above problem?
We have 8 letters out of which A appears twice and B appears three time. Total number of permutation of these letters (without restriction) would be: \frac{8!}{2!3!}=3360.
Now, in half of these cases D will be to the right of C and in half of these cases to the left, hence the final answer would be \frac{3360}{2}=1680
Answer: A.
Hi Bunnel,
If the question were:
Cases in which A is to the right of C?
tot=8!/2!*3!=3360
A to right of C = 3360/(2!)^2?