How many different words can be formed out of the letter of the word “MORADABAD” taken 4 at a time?
A. 620
B. 622
C. 626
D. 624
Answer
641.1k+ views
Hint: In the word $\text{MORADABAD}$ there are two letters that come more than one time. In this word ‘A’ comes thrice and the letter ‘D’ comes twice. So here we will take all the cases in which letters can be formed from the letters of the word $\text{MORADABAD}$.
Complete step by step answer:
Given: The word $\text{MORADABAD}$ is given and we have to find all the possible cases in which $4$ letters word can be formed by this word.
First we will take the word $\text{MORADABAD}$ in which $6$ different letters are used that is: M-$1$ time
O-$1$ time, R-$1$ time, A-$3$ times, B-$1$ time, D-$2$ times.
We have a combination formula that, if we are taking r letters from n letters, then it can be done in ${}^n{C_r}$ ways.
We will find out all the possible cases of word formation.
Case 1st$ \to $ Here we will discuss the word formation cases in which all the letters are different.
So all four letters comes different in ${}^6{C_4} \times 4!\,$ ways
i.e. $\dfrac{{6 \times 5}}{{2!}} \times 4 \times 3 \times 2$
$ = 360$ ways
Case 2nd$ \to $ Here we will discuss in which $2$ alike of one kind and $2$ alike of other kind of letters can come.
So two alike letters of one kind and two alike letters of other kind can come in ${}^2{C_2} \times \dfrac{{4!}}{{2!2!}}$ ways
i.e. $\dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}\, = 6$ ways.
Case 3rd$ \to $Here we will discuss in which two letters are alike and two letters are different.
So two alike and two different letters can come in ${}^2{C_1} \times {}^5{C_2} \times \dfrac{{4!}}{{2!}}$
$ = \dfrac{{2 \times 5 \times 4}}{2} \times \dfrac{{4 \times 5 \times 2 \times 1}}{{2 \times 1}} = 240$ ways.
Case 4th$ \to $ Here we will discuss in which three letters alike and one different come.
So three alike of one kind & one letter different can come in ${}^1{C_1} \times {}^5{C_1} \times \dfrac{{4!}}{{3!}}$ ways
i.e. $5 \times 4 = 20$ ways.
Now we will add all the ways of all the cases.
Hence total possibilities $ = 360 + 6 + 240 + 20$
$ = 626$ ways.
Therefore, 626 different words can be formed out of the letter of the word “MORADABAD” taken 4 at a time. Hence, option (C) is correct.
Note:
In these types of questions, we should always use combination formulas, and keep in mind the repeated letters. Here, as letter ‘A’ comes thrice and letter ‘D’ comes twice and we require $4$ letters out of six different letters so we found all the cases.
Complete step by step answer:
Given: The word $\text{MORADABAD}$ is given and we have to find all the possible cases in which $4$ letters word can be formed by this word.
First we will take the word $\text{MORADABAD}$ in which $6$ different letters are used that is: M-$1$ time
O-$1$ time, R-$1$ time, A-$3$ times, B-$1$ time, D-$2$ times.
We have a combination formula that, if we are taking r letters from n letters, then it can be done in ${}^n{C_r}$ ways.
We will find out all the possible cases of word formation.
Case 1st$ \to $ Here we will discuss the word formation cases in which all the letters are different.
So all four letters comes different in ${}^6{C_4} \times 4!\,$ ways
i.e. $\dfrac{{6 \times 5}}{{2!}} \times 4 \times 3 \times 2$
$ = 360$ ways
Case 2nd$ \to $ Here we will discuss in which $2$ alike of one kind and $2$ alike of other kind of letters can come.
So two alike letters of one kind and two alike letters of other kind can come in ${}^2{C_2} \times \dfrac{{4!}}{{2!2!}}$ ways
i.e. $\dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}}\, = 6$ ways.
Case 3rd$ \to $Here we will discuss in which two letters are alike and two letters are different.
So two alike and two different letters can come in ${}^2{C_1} \times {}^5{C_2} \times \dfrac{{4!}}{{2!}}$
$ = \dfrac{{2 \times 5 \times 4}}{2} \times \dfrac{{4 \times 5 \times 2 \times 1}}{{2 \times 1}} = 240$ ways.
Case 4th$ \to $ Here we will discuss in which three letters alike and one different come.
So three alike of one kind & one letter different can come in ${}^1{C_1} \times {}^5{C_1} \times \dfrac{{4!}}{{3!}}$ ways
i.e. $5 \times 4 = 20$ ways.
Now we will add all the ways of all the cases.
Hence total possibilities $ = 360 + 6 + 240 + 20$
$ = 626$ ways.
Therefore, 626 different words can be formed out of the letter of the word “MORADABAD” taken 4 at a time. Hence, option (C) is correct.
Note:
In these types of questions, we should always use combination formulas, and keep in mind the repeated letters. Here, as letter ‘A’ comes thrice and letter ‘D’ comes twice and we require $4$ letters out of six different letters so we found all the cases.
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