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**Hint:**Here, we will use the selection that all the 4 are different using the combinations

\[{}^n{C_r} = \dfrac{{\left. {\underline {\,

n \,}}\! \right| }}{{\left. {\underline {\,

r \,}}\! \right| \cdot \left. {\underline {\,

{n - r} \,}}\! \right| }}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen. Then simplify to find the required value.

**Complete step-by-step answer:**We are given that the word is COMBINATION.

Since we know that some of the letters are repeated, first we will count the letters in the word 1 C’s, 1 M’s, 1 B’s 1 A’s, 1 T’s, 2 O’s, 2 I’s and 2 N’s.

CASE 1:

When 4 letters are distinct, then we will have

\[ \Rightarrow {}^8{C_4} \times 4!\]

Using the formula for the combinations is \[{}^n{C_r} = \dfrac{{\left. {\underline {\,

n \,}}\! \right| }}{{\left. {\underline {\,

r \,}}\! \right| \cdot \left. {\underline {\,

{n - r} \,}}\! \right| }}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen in the above equation, we get

\[

\Rightarrow \dfrac{{8!}}{{4!\left( {8 - 4} \right)!}} \times 4! \\

\Rightarrow \dfrac{{8!}}{{4!4!}} \times 4! \\

\]

Simplifying the factorials in the above equation, we get

\[

\Rightarrow \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4!4!}} \times 4! \\

\Rightarrow 8 \times 7 \times 6 \times 5 \\

\Rightarrow 1680{\text{ ......eq.(1)}} \\

\]

Thus, there are 1680 ways to make a word, as there 8 different letters in word and letters can be arranged among themselves in \[4!\] ways.

CASE 2:

When 2 letters are repeated, so we will choose the repeated letter from O, I, N in \[{}^3{C_1}\] ways and the remaining two can be selected from 7 in \[{}^7{C_2}\] ways and are arranged among themselves in \[\dfrac{{4!}}{{2!}}\] ways as there are two repeated letters.

So, we have

\[ \Rightarrow {}^3{C_1} \times \dfrac{{4!}}{{2!}} \times {}^7{C_2}\]

Using the formula of the combinations in the above equation, we get

\[

\Rightarrow \dfrac{{3!}}{{1!\left( {3 - 1} \right)!}} \times \dfrac{{4!}}{{2!}} \times \dfrac{{7!}}{{2!\left( {7 - 2} \right)!}} \\

\Rightarrow \dfrac{{3!}}{{1!2!}} \times \dfrac{{4!}}{{2!}} \times \dfrac{{7!}}{{2!5!}} \\

\]

Simplifying the factorials in the above equation, we get

\[

\Rightarrow \dfrac{{3 \times 2!}}{{1!2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2!}} \times \dfrac{{7 \times 6 \times 5!}}{{2 \times 5!}} \\

\Rightarrow \dfrac{3}{1} \times 12 \times \dfrac{{7 \times 6}}{2} \\

\Rightarrow 3 \times 12 \times 21 \\

\Rightarrow 756{\text{ ......eq.(2)}} \\

\]

CASE 3:

When 2 letters of one kind and two alike of another kind, so \[{}^3{C_2}\] ways to find two letter of one kind and other two alike of another kind and are arranged among themselves in \[\dfrac{{4!}}{{2!2!}}\] ways as there are two repeated letters.

So, we have

\[ \Rightarrow {}^3{C_2} \times \dfrac{{4!}}{{2!}}\]

Using the formula of the combinations in the above equation, we get

\[

\Rightarrow \dfrac{{3!}}{{2!\left( {3 - 2} \right)!}} \times \dfrac{{4!}}{{2! \times 2!}} \\

\Rightarrow \dfrac{{3!}}{{2!1!}} \times \dfrac{{4!}}{{2! \times 2!}} \\

\]

Simplifying the factorials in the above equation, we get

\[

\Rightarrow \dfrac{{3 \times 2!}}{{1!2!}} \times \dfrac{{4 \times 3 \times 2!}}{{2 \times 2!}} \\

\Rightarrow \dfrac{3}{1} \times \dfrac{{4 \times 3}}{2} \\

\Rightarrow 3 \times 6 \\

\Rightarrow 18{\text{ ......eq.(3)}} \\

\]

Adding the equation (1), equation (2) and equation (3) to find required value, we get

\[

\Rightarrow 1680 + 756 + 18 \\

\Rightarrow 2454{\text{ ways}} \\

\]

**Hence, the option B is correct.**

**Note:**Some students use the formula of permutation, \[{}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\] instead of combinations is \[{}^n{C_r} = \dfrac{{\left. {\underline {\,

n \,}}\! \right| }}{{\left. {\underline {\,

r \,}}\! \right| \cdot \left. {\underline {\,

{n - r} \,}}\! \right| }}\], where \[n\] is the number of items, and \[r\] represents the number of items being chosen, which is wrong. If we go with , then we will end up considering the similar letters as distinct letters and will get more than the possible cases. Since we know that permutation is the number of arrangements of all those elements that have been chosen in the time of combination, we say \[{}^n{P_r} = r!{}^n{C_r}\].

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