Question

In how many ways 7 letters a.,b, c, d, e, f, g, can be arranged in wch c and e can't together

Answer 1

First arrange these seven letters without any constraints. So the total number of ways to do this is 7!, as we have total seven letters.

Now according to given situation we have to make words which doesn't have C&E together. So solve this let's suppose that C&E as one unit, now total number of words become 6!, as we have left with 6 letters.

Now C&E are interchangeable so now again we have 6! ways to arrange them.

So the required answer is

=7!-6!-6!

=7*6*5*4*3*2*1–2(6*5*4*3*2*1)

=5040–2(720)

=5040–1440

=3600

So there are 3600 words which can be made according to given condition.

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Answer 2

We have total of 7 letters, which can be arranged in 7! ways.

Now assume that CE is a single letter. So that we now have 6 letters. we can arrange these 6 letters in 6! ways. Now C & E can be arranged as CE & EC, so end up having 2 x 6! arrangements where C &E appear together.

Total possible arrangement : 7!

Arrangements where C & E are together: 2x6!

So, arrangements where C&E are not together:

7! - 2x6! = 7x6! -2x6!

= 5x6! = 5x 720= 3600. Solved!