In how many ways can the 26 letters of the English alphabet be arrange

A & B have to be separated by 7 letters

Make a block of 9 letters with {A/B, _, _, _, _, _, _, _, B/A} and remaining 17 letters

We can arrange 7 letters from 24 available in \(24p_7\) ways

A & B can be interchanged in 2 ways

18 units (1 block + remaining 17 letters) can be arranged in 18! ways

--> Total number of ways = 2*24p7*18! = 2*{24*23*22*21*20*19*18}*18! = 2*24!*18 = 36*24!

Option B

TOTAL 26 LETTERS

A__[7]__B___[17]___
The number ways to choose [7] in between is 24C7
The number of arrangements of [7] is 7!
The number of arrangements of A and B is 2
The number of arrangements of [17] and "A_[7]_B" as a group is 18!
Total arrangements 24!/7!17! * 7! * 2 * 18! = 36*24!

Ans (B)

If we take out A and B from the 26 letters, we are left with 24 letters.
The number of ways that 7 letters can be selected out of 24 = 24C7.
The number of ways each of the 7 letters selected can be arranged = 7!
After selecting 7 letters out of 24, the number of letters left = 24-7=17
In how many ways can we arrange the remaining 17 letters? 17!
Each of the nine letters formed by interspersing 7 letters between letters A and B can be arranged between the 17 remaining letters in 17+1=18 ways
Last but not the least, the two letters A and B can be arranged interchangeably at the ends in 2 ways

Total No. of ways = \(24C7*7!*17!*18*2\) = \((24!/(7!*17!))\)\(*7!*17!*18*2\) = \(36*24!\)

The answer is B.

7 letters must separate A and B.

Implication:
In the 26-letter arrangement, A and B must occupy the ends of a 9-letter block, as in the following examples:
A=1 __ __ __ __ __ __ __ B=9
A=2 __ __ __ __ __ __ __ A=10
A=17 __ __ __ __ __ __ __ B=25
A=18 __ __ __ __ __ __ __ B=26

The value in red indicates that the number of ways to position this 9-letter block = 18
Number of ways to arrange A and B at the ends of this block = 2!
Number of ways to arrange the remaining 24 letters = 24!
To combine these options, we multiply:
18 * 2! * 24! = 36 * 24!

In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B?

A. 12

B. 36∗24!

C. 24C7∗18!∗2

D. 26C7×20!×2

E. 26P7×20!×2

Assumption: Arrangements is linear not circular.

A + seven alphabets + B = 9 alphabets.
A and B can be written in 2 ways:
1. AB
2. BA

Rest of the places can be filled with any random alphabets that are left i.e. 24 alphabets.
A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ (each space is an alphabet)
_ A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
_ _ A_ _ _ _ _ _ _ B _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
...
...
_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ A _ _ _ _ _ _ _ B

Since either of A or B can take place till 9th last alphabet i.e. 18th position,
total positions that A or B can take = 18

So, Total ways of arrangements =
24! * 18 * 2 = 24! * 36

Answer B.

26 - 9 = 17 ... If we consider the 9 letters as 1 objects we have to permute 18 different objects .. b and A can interchange ... I think the answer is C

A _ _ _ _ _ _ _ B(24C7, as A and B are fixed) + remaining letters(18!)
+B _ _ _ _ _ _ _ A (24C7)+ remaining letters(18!)
--------------------------------------------------
=> 24C7 * 18! * 2
Ans.C

In how many ways can the 26 letters of the English alphabet be arranged so that there are seven letters between the letters A and B?

The position of A, B and the other 7 letters can be fixed in the following ways :
A 24C7 B
B 24C7 A
So, there can be 2 * 24C7 arrangements for these 9 letters. Now we can say these 9 letters are serving as a single letter. So the total letters become 18. These 18 letters can be arranged in 18! ways. So, the total arrangements will become : 2* 24C7 * 18!.
C is the answer.

Arranging these 7 numbers between A & B = 24*23*.....18
Number of ways of arranging A & B = 2!

Considering all the 9 leeters as one arrangement
Number of ways of arranging the 18 letters = 18!

Therefore the tolal way of arranging the letters equal= 24*23....18*18!*2
=>24! *2*18 = 36*24
Therefore IMO B

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