Question

If number of ways in which 8 letters A,B,C,D,E,F,G,H are arranged in a row such that A,B,C,D are in alphabetical order but A,B,C,D,E are not in alphabetical order is N, then the value of (1350 - N)is ?

Answer 1

Hint: Here we will use permutation to find number of ways in which all $8$ letters can be arranged and then subtract number of ways in which $A,B,C,D$ are in alphabetical order but $A,B,C,D,E$ are not in alphabetical order. In one case we take $(A,B,C,D)$ as one letter and $E,F,G,H$ as separate letters, in other case we take $(A,B,C,D,E)$ as one letter and $F,G,H$ as separate letters.

Complete step-by-step answer:
Given, $8$ letters $A,B,C,D,E,F,G,H$
To find number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D$ are always in alphabetical order, we consider $(A,B,C,D)$ as one letter and $E,F,G,H$ as other $4$ different letters.
Therefore, we have total $5$ letters that have to be arranged.
We use the formula of permutation to find number of possible ways of arrangement of $n$ letters
ie; If number of letters $= n$ , then number of possible ways to arrange $n$ letters $= n!$
where; $n! = n \times (n - 1) \times (n - 2) \times ............. \times 3 \times 2 \times 1$
So, number of ways in which $5$ letters (where $(A,B,C,D)$ is one letter) can be arranged $= 5!$
Ie; $5! = (5) \times (4) \times (3) \times (2) \times (1) = 120$
Thus, there are $120$ ways to arrange 8 letters $A,B,C,D,E,F,G,H$ such that $A,B,C,D$ are always in alphabetical order. $.........(i)$
Now, to find number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D,E$ are always in alphabetical order, we consider $(A,B,C,D,E)$ as one letter and $F,G,H$ as other $3$ different letters.
Therefore, we have total $4$ letters that have to be arranged.
We use the formula of permutation to find number of possible ways of arrangement of $n$ letters
ie; If number of letters $= n$ , then number of possible ways to arrange $n$ letters $= n!$
where; $n! = n \times (n - 1) \times (n - 2) \times ............. \times 3 \times 2 \times 1$
So, number of ways in which $4$ letters (where $(A,B,C,D,E)$ is one letter) can be arranged $= 4!$
Ie; $4! = (4) \times (3) \times (2) \times (1) = 24$
Thus, there are $24$ ways to arrange 8 letters $A,B,C,D,E,F,G,H$ such that $A,B,C,D,E$ are always in alphabetical order. $.........(ii)$
Thus, number of ways in which $8$ letters $A,B,C,D,E,F,G,H$ are arranged in a row such that $A,B,C,D$ are in alphabetical order but $A,B,C,D,E$ are not in alphabetical order $= N$
Ie; Number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D$ are always in alphabetical order $-$ number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D,E$ are always in alphabetical order $= N$
Ie; $N = equation(i) - equation(ii)$
Ie; $N = 120 - 24$
Ie; $N = 96$ $.........(iii)$
Now, to find the value of $(1350 - N)$ , we substitute the value of $N = 96$ from $equation(iii)$ ;
Ie; $(1350 - N) = (1350 - 96) = 1254$
Therefore; the value of $(1350 - N) = 1254$

Note: In these types of questions, we always try to pair up the letters that have to be in alphabetical order and then we add or subtract according to the requirement.
Alternative method: $N =$ number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D$ are always in alphabetical order $-$ number of ways in which 8 letters $A,B,C,D,E,F,G,H$ are arranged such that $A,B,C,D,E$ are always in alphabetical order i.e. $N = 5! - 4! = (5 \times 4 \times 3 \times 2 \times 1) - (4 \times 3 \times 2 \times 1) = 120 - 24 = 96$