Suppose A = 1 , B = 2 , C = 3 D = 4 ... X = 24 , Y = 25 Z = 26

$$\frac{(A-M)(D-M)(G-M)\cdots(S-M)}{(M-Z)(M-Y)(M-X)\cdots(M-T)}$$​​

Correct option is A

Given:
A = 1, B = 2, C = 3, ..., M = 13, ..., X = 24, Y = 25, Z = 26
​$$\frac{(A-M)(D-M)(G-M)\cdots(S-M)}{(M-Z)(M-Y)(M-X)\cdots(M-T)}$$​​
Expression:
Numerator: (A − M)(D − M)(G − M) ... (M − M) ... (S − M)
Denominator: (M − Z)(M − Y)(M − X) ... (M − T)

Solution:
​In the numerator, one of the terms is (M − M) which equals 0.

Any product that includes 0 will result in the entire numerator being 0, regardless of the other terms.

So the whole expression becomes:

0 ÷ (something non-zero) = 0

Final Answer: (A) 0