Question

The 60 permutations of ALLEN are arranged in dictionary order, as if each were on ordinary five letter words. The last letter of word which occurs at ${23^{{\text{rd}}}}$ position is:-
A.A
B.E
C.L
D.N

Answer 1

Hint: The total number of different ways that $n$ items can be arranged is $n!$, provided that no item is repeated. If any item is repeated $m$times, then the total possible arrangements are then reduced by a factor of $m!$.

Complete step-by-step answer:
For an sequence of $n$ items, with exactly one item repeating $m$ times, the total number of possible different ways that these $n$ items can be arranged is given by $\dfrac{{n!}}{{m!}}$.
For the word ALLEN, there is one repeating alphabet, that is L is repeated 2 times.
The total number of possible different ways the alphabets of the word ALLEN can be arranged is $\dfrac{{5!}}{{2!}} = 60$
When these words are arranged in the dictionary the words starting with A will be coming first.
The total number of words present starting with A can be found by finding the ways the four alphabets L,L,E and N can be arranged.
Thus words starting with A are
$\dfrac{{4!}}{{2!}} = 12$
Thus the first 12 words will be starting with A.
The next words will be starting from E.
Similarly, words starting with E will be
$\dfrac{{4!}}{{2!}} = 12$
Thus the 23rd word will be starting with the alphabet E.
Thus the problem is reduced to find the $23 - 12 =$11th word with four alphabets A,L,L and N.
The words starting with A will be $\dfrac{{3!}}{{2!}} = 3$.
Similarly, words starting with alphabet L will be $3! = 6$
Words starting with alphabet N will be $\dfrac{{3!}}{{2!}} = 3$
Thus the second alphabet of the 23rd word will be N.
The problem is reduced to finding the $11 - 3 - 6 =$2 and number from the alphabets A,L and L.
The three possible combinations are ALL, LAL and LLA.
The second combination is LAL.
Thus the 23rd word becomes ENLAL.
The last letter of the word which occurs at 23rd position is L.

Thus the option C is the correct answer.

Note: If there are multiple repeating items in the sequence of $n$ items, the contribution of each repeating element is divided separately. For two repeating items ${m_1}$and ${m_2}$ times, the total number of possible ways of arrangement becomes $\dfrac{{n!}}{{{m_1}!{m_2}!}}$.