Problem Solving

Gmat_mission wrote: ↑

Sun Dec 19, 2021 1:23 am

The letters \(C, I, R, C, L,\) and \(E\) can be used to form \(6\)-letter strings such as \(CIRCLE\) or \(CCIRLE.\) Using these letters, how many different \(6\)-letter strings can be formed in which the two occurrences of the letter \(C\) are separated by at least one other letter?

A. 96
B. 120
C. 144
D. 180
E. 240

Answer: E

Source: Official Guide

We can use the rule that says: TOTAL number of outcomes if we IGNORE the rule = (number of outcomes that FOLLOW the rule) + (number of outcomes that BREAK the rule)
In other words: Number of ways to arrange the 6 letter if we IGNORE the rule = (number of words that DON'T have adjacent C's) + (number of words that DO have adjacent C's)

Rearrange to get: number of words that DON'T have adjacent C's = (Number of ways to arrange the 6 letter if we IGNORE the rule) - (number of words that DO have adjacent C's)

Number of ways to arrange the 6 letter if we IGNORE the rule
If we IGNORE the rule, then we are arranging the letters in CIRCLE
Since we have DUPLICATE letters, we can apply the MISSISSIPPI rule (see video below)

In the word CIRCLE:
There are 6 letters in total
There are 2 identical C's
So, the total number of possible arrangements = 6!/(2!) = 360

number of words that DO have adjacent C's
Take the two C's and "glue" them together to get the SUPER LETTER "CC"
This ensures that the C's are together
We now must arrange CC, I, R, L, E
We can arrange n different objects in n! ways
So, we can arrange CC, I, R, L, and E in 5! ways (= 120 ways)
So, number of words that DO have adjacent C's = 120

So, number of words that DON'T have adjacent C's = 360 - 120 = 240

Answer: E