Question

The number of letters that can be made by taking four letters of the word COMBINATION is-
$\left( a \right)70$
$\left( b \right)63$
$\left( c \right)3$
$\left( d \right)136$

Answer 1

We need to understand that there are 3 ways of selecting four letters of the word COMBINATION
With the help of which we can find out the number of letters.
This can be made by taking four letters from the word.

Formula used:
$_{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$, where n represents the total number of items out of which selection would be made and r represents the few items which have to be chosen.

Complete step-by-step answer:
It is given that the word COMBINATION contains 11 letters
Out of which 3 letters are present two times and 8 is present only one time.
Therefore we have to find four letters that can be selected in 3 ways.
First way: Four letters can be different that is,
$_{}^8{C_4}$ $= \dfrac{{8!}}{{4!(8 - 4)!}}$
On subtracting the bracket term we get,
$= \dfrac{{8!}}{{4!(4)!}}$
We can split it as,
$= \dfrac{{8 \times 7 \times 6 \times 5 \times 4!}}{{4! \times 4 \times 3 \times 2 \times 1}}$
On simplifying, we get
$= 70$
Second way: Two letters can be similar and two are different, that is, $_{}^3C_1^{}$and
The two different letters will have to be selected from the remaining 7 letters, that is, $_{}^7{C_2}$.
So we can write $_{}^3C_1^{} \times _{}^7{C_2} = \dfrac{{3!}}{{1!(3 - 1)!}} \times \dfrac{{7!}}{{2!(7 - 2)!}}$
On subtract the terms,
$= \dfrac{{3!}}{{1!(2)!}} \times \dfrac{{7!}}{{2!(5)!}}$
we can split it as,
$= \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{7 \times 6 \times 5!}}{{2 \times 1 \times 5!}}$
On some simplifying,
$= 63$
Third way: Four letters can be chosen from 3 letters which are present two times, that is, $_{}^3{C_2}$.$= \dfrac{{3!}}{{2!(3 - 2)!}}$
$= \dfrac{{3 \times 2!}}{{2!}} = 3$
Therefore, total number of ways by which 4 letters can be chosen$= _{}^8{C_4} + _{}^3{C_1} \times _{}^7{C_2} + _{}^3C_2^{}$ $= 70 + 63 + 3 = 136$
Thus, the total number of ways by which we can four letters from the word COMBINATION is 136

So, the correct answer is “Option d”.

Note: Combination in mathematics is regarded as the chance or possibility of selection of few items from a huge number of items.
The number of combinations of $n$ different things taken $r$ at a time allowing repetitions is $_{}^{n + r - 1}{C_r}$
The number of selections of $r$ objects out of $n$ identical objects is $1$