Question

How many arrangements of INCONSISTENT are there in which NE appear consecutively or NO appear consecutively but not both NE and NO are consecutive?

The answer is 2x11!/2!2!2!2! - 2x10!/2!2!2!

Please show me how to get the answer, it's important for my midterm, thanks.

Answer 1

There are 12 letters in the word INCONSISTENT. Fix NE as one letter. Then there will be 11 letters. Out of the 11 letters there are 2 repetitions of 4 letters (N,I S and T repeats each 2 times). The repetitions has to be divided from the total arrangements Hence 11 are arranged in 11!/2!2!2!2! ways. Same is the case with NO. Treat NO as 1 letter. Hence the same number of arrangement for NO.

Now consider both NO and NE each as one letter. There will be 10 letters in total so both are consecutive in 10!/2!2!2! ways

Let n(NE) represents the number in which NE is consecutive, n(NE) has 10!/2!2!2! arrangements where NO is consecutive

and Let n(NO) represents the number in which NO is consecutive, n(NO) has again 10!/2!2!2! arrangements where NE is consecutive

So we need to subtract 2* 10!/2!2!2! from the total number of ways

The total number of ways is n(NE)+n(NO)=11!/2!2!2!2!+11!/2!2!2!2!=2*11!/2!2!2!2!

Hence the required number of ways is 2*11!/2!2!2!2!-2*10!/2!2!2!