Question

The number of ways in which any four letters can be selected from the word ‘CORGOO’ is?
A. $15$
B. $11$
C. $7$
D. None of these

Answer 1

The given question involves the concepts of permutations and combinations. We are required to find the number of ways of selecting four letters using alphabets of word “CORGOO”. First we count the number of different letters in the word “CORGOO”, 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 four letters are distinct, the second case is when two O’s are selected, the third case is when three O’s are selected. We find the number of possibilities in every 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 CORGOO. We are to find a number of ways in which $4$ letters are to be chosen from the letters in CORGOO. Here, some of the letters are repeated in the word so we can’t simply use ${}^6{C_4}$. So, let's first count the letters in the word.
Number of O’s in the word ‘CORGOO’ $= 3$
Number of C’s in the word ‘CORGOO’ $= 1$
Number of R’s in the word ‘CORGOO’ $= 1$
Number of G’s in the word ‘CORGOO’ $= 1$
Since the letters of the word are being repeated. So, we will form cases.

Case $1$: When all the $3$ O’s are selected.
Since three O’s are selected.
So, we have to choose one letter out of the three different letters from the word ‘CORGOO’. So, the number of ways of doing so is $^3{C_1}$.
Now, using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$^3{C_1} = \dfrac{{3!}}{{1! \times 2!}}$
Cancelling the common factors in numerator and denominator, we get,
${ \Rightarrow ^3}{C_1} = \dfrac{6}{2} = 3$

Case $2$: When two O’s are selected.
Since two O’s are selected.
So, we have to choose two letters out of the three different letters from the word ‘CORGOO’. So, the number of ways of doing so is $^3{C_2}$.
Now, using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$^3{C_2} = \dfrac{{3!}}{{1! \times 2!}}$
Cancelling the common factors in numerator and denominator, we get,
${ \Rightarrow ^3}{C_2} = \dfrac{6}{2} = 3$

Case $2$: When only one O is selected.
Since one O is selected.
So, we have to choose three letters out of the three different letters from the word ‘CORGOO’. So, the number of ways of doing so is $^3{C_3}$.
Now, using the formula, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, we get,
$^3{C_3} = \dfrac{{3!}}{{0! \times 3!}}$
We know that the value of $0!$ is $1$.
Cancelling the common factors in numerator and denominator, we get,
${ \Rightarrow ^3}{C_3} = 1$
So, the total number of ways in which any four letters can be selected from the word ‘CORGOO’ is $3 + 3 + 1 = 7$.

So, option C is the correct answer.

Note: A combination is calculating the number of ways in which ‘r’ things are to be selected out of ‘n’ total different things. The formula for combination is given by ${}^n{C_r}$, and this expression is equal to, ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. If we try to solve the problem with ${}^6{P_4}$, then we end up considering the alike letters as distinct letters and will get more than the possible cases and end up with wrong answers for the problem. We must take care of calculations in such questions.