Answer

Verified

433.8k+ views

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.

Recently Updated Pages

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

Trending doubts

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

What is pollution? How many types of pollution? Define it