# What are the total number of 6-digit numbers that can be made with the digits 1,2,3 and 4 having exactly two pairs of alike digits.

$

{\text{A}}{\text{. 480}} \\

{\text{B}}{\text{. 540}} \\

{\text{C}}{\text{. 1080}} \\

$

${\text{D}}{\text{. }}$None of these

Last updated date: 28th Mar 2023

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Answer

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Hint: Here, we will proceed by multiplying the number of ways of selecting two digits for two pairs of alike digits and the number of ways of selecting the remaining two digits for two different digits. The, we will further multiply the product obtained with the number of ways in which these 6-digit numbers can be arranged.

Complete step-by-step answer:

We have to form 6-digit numbers using digits 1,2,3 and 4 which will have exactly two pairs of alike digits.

A 6-digit number formed will have two alike pairs of digits (i.e., total four digits) and two different digits from those which are already used.

Total number of digits out of which digits (1,2,3,4) for the required 6-digit numbers is selected = 4.

Since, the number of ways of selecting r items out of n items is given by ${}^n{C_r}$.

The number of ways of selecting two digits for two pairs of alike digits out of total four digits is ${}^4{C_2}$ and the number of ways of selecting two different digits out of the two digits left is ${}^2{C_2}$.

As we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

So, the number of ways of selecting two digits for two pairs of alike digits and two different digits out of the digits 1,2,3,4

$

= {}^4{C_2} \times {}^2{C_2} \\

= \left[ {\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}} \right] \times \left[ {\dfrac{{2!}}{{2!\left( {2 - 2} \right)!}}} \right] \\

= \left[ {\dfrac{{4.3.2!}}{{2.1.2!}}} \right] \times \left[ {\dfrac{{2!}}{{2!0!}}} \right] \\

= \left[ {\dfrac{{4 \times 3}}{{2 \times 1}}} \right] \times \left[ {\dfrac{1}{{0!}}} \right] \\

= 6 \times 1 \\

= 6 \\

$

Since, the number of ways of arranging n items out of which x items of one type are repeating and y items of another type are repeating is given by $\dfrac{{n!}}{{x!y!}}$.

Also, the number of ways in which these 6-digit numbers which consists of two pairs of alike digits and two different digits can be arranged

\[ = \dfrac{{6!}}{{2!2!}} = \dfrac{{6.5.4.3.2!}}{{2.1.2!}} = \dfrac{{6.5.4.3}}{2} = 180\]

Therefore, required total number of 6-digits formed = (number of ways of selecting two digits for two pairs of alike digits and two different digits out of the digits 1,2,3,4)$ \times $( number of ways in which these 6-digit numbers which consists of two pairs of alike digits and two different digits can be arranged)

$ \Rightarrow $Required total number of 6-digits formed$ = 6 \times 180 = 1080$

Hence, option C is correct.

Note: In this particular problem, the 6-digit numbers which are to be obtained must contain both two pairs of alike digits and two different digits apart from already selected digits that’s why we are multiplying ${}^4{C_2}$ and ${}^2{C_2}$ with each other. Also, it is important to note that for selection purposes we use combinations and for arranging purposes we use permutations.

Complete step-by-step answer:

We have to form 6-digit numbers using digits 1,2,3 and 4 which will have exactly two pairs of alike digits.

A 6-digit number formed will have two alike pairs of digits (i.e., total four digits) and two different digits from those which are already used.

Total number of digits out of which digits (1,2,3,4) for the required 6-digit numbers is selected = 4.

Since, the number of ways of selecting r items out of n items is given by ${}^n{C_r}$.

The number of ways of selecting two digits for two pairs of alike digits out of total four digits is ${}^4{C_2}$ and the number of ways of selecting two different digits out of the two digits left is ${}^2{C_2}$.

As we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$

So, the number of ways of selecting two digits for two pairs of alike digits and two different digits out of the digits 1,2,3,4

$

= {}^4{C_2} \times {}^2{C_2} \\

= \left[ {\dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}} \right] \times \left[ {\dfrac{{2!}}{{2!\left( {2 - 2} \right)!}}} \right] \\

= \left[ {\dfrac{{4.3.2!}}{{2.1.2!}}} \right] \times \left[ {\dfrac{{2!}}{{2!0!}}} \right] \\

= \left[ {\dfrac{{4 \times 3}}{{2 \times 1}}} \right] \times \left[ {\dfrac{1}{{0!}}} \right] \\

= 6 \times 1 \\

= 6 \\

$

Since, the number of ways of arranging n items out of which x items of one type are repeating and y items of another type are repeating is given by $\dfrac{{n!}}{{x!y!}}$.

Also, the number of ways in which these 6-digit numbers which consists of two pairs of alike digits and two different digits can be arranged

\[ = \dfrac{{6!}}{{2!2!}} = \dfrac{{6.5.4.3.2!}}{{2.1.2!}} = \dfrac{{6.5.4.3}}{2} = 180\]

Therefore, required total number of 6-digits formed = (number of ways of selecting two digits for two pairs of alike digits and two different digits out of the digits 1,2,3,4)$ \times $( number of ways in which these 6-digit numbers which consists of two pairs of alike digits and two different digits can be arranged)

$ \Rightarrow $Required total number of 6-digits formed$ = 6 \times 180 = 1080$

Hence, option C is correct.

Note: In this particular problem, the 6-digit numbers which are to be obtained must contain both two pairs of alike digits and two different digits apart from already selected digits that’s why we are multiplying ${}^4{C_2}$ and ${}^2{C_2}$ with each other. Also, it is important to note that for selection purposes we use combinations and for arranging purposes we use permutations.

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