Animated Solution for Mathematics - Permutations and Combinations: From all the English alphabets, five letters are chosen and are arranged in alphabetical order. The total number of ways, in which the middle letter is 'M', is:

Imagine standing before the 26 letters of the English alphabet. You are tasked with a challenge: select five letters and arrange them in alphabetical order, with the strict condition that the middle letter must be 'M'.

At first glance, this might seem like a complex permutation problem, but let us peel back the layers and reveal the elegant simplicity hidden within.

The most important realization in this problem is the nature of the 'alphabetical order' constraint. Many students instinctively reach for the factorial operator, thinking, 'I have five letters, so I must arrange them in $5!$ ways.'

But pause for a moment. If you select the set of letters $\{A,C,E,M,Z\}$, how many ways can you arrange them in alphabetical order? Only one: $A,C,E,M,Z$.

The constraint of alphabetical order is not a hurdle; it is a gift. It means that once you have selected your set of letters, the arrangement is already determined. We do not need to arrange them; we only need to select them.

The problem tells us that the middle letter is fixed as 'M'. This is our anchor.

Because the sequence must be alphabetical, any letter placed in the first or second position must come from the set of letters that appear before 'M' in the alphabet. Similarly, any letter in the fourth or fifth position must come from the set of letters that appear after 'M'.

Let us count them: The letters before 'M' are $\{A,B,C,D,E,F,G,H,I,J,K,L\}$, which gives us a pool of 12 letters. The letters after 'M' are $\{N,O,P,Q,R,S,T,U,V,W,X,Y,Z\}$, which gives us a pool of 13 letters.

Now, the problem transforms into a simple selection task. We need to choose 2 letters from the 12 available in the pre-'M' pool and 2 letters from the 13 available in the post-'M' pool.

The number of ways to choose 2 letters from 12 is given by the combination formula ${}^{12}{C}_2$:

Similarly, the number of ways to choose 2 letters from 13 is ${}^{13}{C}_2$:

According to the Fundamental Principle of Counting, if one task can be done in $m$ ways and another in $n$ ways, the total number of ways to perform both is $m\times n$.

Here, we have 66 ways to choose the first two letters and 78 ways to choose the last two letters. Multiplying these together, we get:

There you have it! The complexity vanishes when you understand the constraints. You have successfully navigated the trap of unnecessary permutations and arrived at the elegant solution of 5148.

Keep this mindset—always look for the constraint that simplifies the problem rather than complicates it.