Counting

Question

A number of ways we can arrange letters of the word " TESTBOOK " such that E always comes between O's is ___________

Answer 1

We have 8 letters, with O and T repeating twice.

We need to find the number of ways of arranging the word "TESTBOOK" such that E always comes between O's

Essentially, we need to group OEO. So we have T T OEO B K S

We can arrange them in $6!$ ways. But since we have repeating T's and O's. We divide by $2! \times 2!$

$\therefore$ Possible arrangements $= \frac{6!}{2! \times 2!}$

 

Edit:

As @minipanda mentioned, it is not necessary that E should always be in group.

Therefore, out of 8 blocks, we choose 3 such that O $\rightarrow$ E $\rightarrow$ O

And then arrange the remaining 5

we get $\binom{8}{3} \times \frac{5!}{2!} = 3360$

Comments

Changed

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Another approach I thought is like :

Total number of arrangements : 8!/(2!*2!)

But these arrangements include orders like O → E → O , E → O → O , O → O → E. But we only want the first one instead of 3. 

That is in place of 3 orders we want only 1. So divide  8!/(2!*2!) by 3. We get 3360.

More details :

Let x1,x2,x3,x4 be strings of letters comprising of different combinations of {T,T,S,B,K}.

For instance let x1=S,x2=B,x3=T,x4=KT. Then

x1 O x2 E x3 O x4

x1 E x2 O x3 O x4

x1 O x2 O x3 E x4

these are the arrangements we get out of which we want only one.

For each such x1,x2,x3,x4 we will get 3 orders and we want 1. So divide the whole no. of possible arrangement with 3.

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thanks Minipanda