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.

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

Depends. What's the couple's last name? (Or are you saying the child will not share the parents' last name?)

  • $\begingroup$ Wow. That was brilliant. Their last name was Zeta, but I didn't think it mattered until you said it. $\endgroup$ – user230452 Feb 25 '16 at 5:27
  • $\begingroup$ Now, I see why it is $\binom{25}{2}$. $\endgroup$ – user230452 Feb 25 '16 at 5:28

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