# How many possible monograms are there?

The question is simple. A couple want to name their baby in such a way that his initials are in alphabetical order, with no repetition.The monogram is of the form ABC, where A is the initial of his first name, B his second name and C his last name.

My reasoning is simple. I realised that it's equivalent to picking up $3$ elements from a list of $26$ elements. We don't have to worry about counting if they're in alphabetical order because every $3$ tuple chosen like this will have one and only one alphabetical order. In other words, there is a one to one correspondence in between the number of ways of choosing $3$elements and the number of ways of arranging these $3$ elements in alphabetical order.

My answer is $$\binom{26}{3}$$

However, it is wrong. What is the mistake here ?

Edit : I left out the name of the couple because I didn't think it mattered. Their name was Mr. And Mrs. Zeta. Now, I see why the answer makes sense. The baby shares the parents last name and the remaining two alphabets can be chosen in $\binom{25}{2}$ ways, each of which correspond with only one alphabetical arrangement since they are all different combinations.

• The answer is indeed $\binom{26}{3}$. Clearly described reasoning. – André Nicolas Feb 25 '16 at 5:21
• Why do you claim it is wrong? – Eric Towers Feb 25 '16 at 5:23
• The book gives the answer as $\binom{25}{2}$ – user230452 Feb 25 '16 at 5:26

• Now, I see why it is $\binom{25}{2}$. – user230452 Feb 25 '16 at 5:28