Answer
Verified
421.5k+ 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
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE