Question

The number of 4 letter words that can be formed using the letters of the word COMBINATION is
a. 2436
b. 2454
c. 1698
d. 498

Answer 1

Hint: First we count the number of different letters in the word COMBINATION, then we will make cases if the letters are repeated or not then by using the formula of combination i.e. \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], the cases will be that all the 4 letters are distinct, the other case is when 2 letters are same and the other 2 are distinct, and the third case is when all the 4 letters are from the repeated letters. We find the number of possibilities in that particular case. At last, we’ll add the result of all the cases to get the answer.

Complete step by step Answer:

We are given the word, COMBINATION,
We are to find 4 letter words using the letters in COMBINATION.
Hence some of the letters are repeated so we can’t go simply go \[{}^{11}{P_4}\],
So, lets first count the letters in the word, i.e.,
C, M, B, A, T - 1
O, I, N – 2
When 4 letters are distinct, we have
\[ = {}^8{C_4} \times 4!\]
Now using the fact, \[{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\], we get,
\[ = \dfrac{{8!}}{{4!4!}} \times 4!\]
\[ = \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!}}\]
\[ = 1680\] ways to make a word, as there are 8 different letters in the word and letters can be arranged among themselves in 4! ways.
When 2 letters are repeated, we can choose the repeated letter from O, I, N in \[{}^3{C_1}\] ways and the remaining two can be selected from the other 7 in \[{}^7{C_2}\] ways and they can be arranged among themselves in \[\dfrac{{4!}}{{2!}}\]ways as there are two repeated letters.
So, we have,
\[
  {}^3{C_1} \times \dfrac{{4!}}{{2!}} \times {}^7{C_2} \\
   = \dfrac{{3!}}{{2! \times 1!}} \times \dfrac{{4!}}{{2!}} \times \dfrac{{7!}}{{2!5!}} \\
   = \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2!}} \times \dfrac{{7 \times 6 \times 5!}}{{2!5!}} \\
   = 3 \times 4 \times 3 \times 7 \times 3 \\
   = 756{\text{ }}ways \\
 \]
Now, when two letters are of one kind and two alike of another kind,
We get,\[{}^3{C_2}\]ways to find two letters of one kind and other two alike of another kind. And they can be arranged among themselves in \[\dfrac{{4!}}{{2!2!}}\]ways as there is two repeated letter.
So, we get,
\[
  {}^3{C_2} \times \dfrac{{4!}}{{2!2!}} \\
   = \dfrac{{3!}}{{1!2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2!2!}} \\
   = 3 \times 6 \\
   = 18{\text{ }}ways \\
 \]
Now, the total number of ways,
\[1680 + 756 + 18 = 2454\] ways.
Hence, the correct option is (b).

Note: If we go with \[{}^{11}{P_4}\],then we end up considering the similar letters as distinct letters and will get more than the possible cases, as for example “N” and “N” are same, so if a word is formed “NANC” and “NANC” they will be the same, as both the N’s are same.
 A permutation is selecting all the ordered pair of ‘r’ elements out of ‘n’ total elements is given by ${}^n{P_r}$, and this expression is equal to
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$
It can also be said for arranging all the elements in order after selecting combinations of ‘r’ element out of total ‘n’ elements, where expression for combination is ${}^n{C_r}$, and this expression is equal to

Since we said that permutation is the number of arrangements of all those elements that have been chosen in the time of combination, we say that
${}^n{P_r} = r!{}^n{C_r}$
Or for more simplification, we can conclude that
$
  {}^n{P_r} = r!\dfrac{{n!}}{{r!\left( {n - r} \right)!}} \\
   \Rightarrow {}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}} \\
 $