Hi, could someone provide a more detailed answer for the following permutation question:
Q. How many six-letter sequences consisting of 1A, 2Bs and 3Cs are possible?
The answer guide gave the following:
If all letters in sequence different, 6! sequences possible.
But, arrangement of 2Bs and 3Cs need to be removed, therefore ans = 6! /((2!)(3!))= 60.
I'm having trouble understanding why, to remove the arrangement of 2Bs and 3Cs, the division needs to occur. Thanks. [/url]
Q. How many six-letter sequences consisting of 1A, 2Bs and 3Cs are possible?
The answer guide gave the following:
If all letters in sequence different, 6! sequences possible.
But, arrangement of 2Bs and 3Cs need to be removed, therefore ans = 6! /((2!)(3!))= 60.
I'm having trouble understanding why, to remove the arrangement of 2Bs and 3Cs, the division needs to occur. Thanks. [/url]
