Number of 6 digit numbers that can be made with the digits 1,2,3 and 4 and having exactly two pairs of digits is

Question

Number of 6 digit numbers that can be made with the digits 1,2,3 and 4 and having exactly two pairs of digits is (A) 978 (B) 1801 (C) 1080 (D) 789

Tardigrade Admin · Accepted Answer

The number will have 2 pairs and 2 different digits. The number of selections = 4 C 2 × 2 C 2, and for each selection, number of arrangements =(6 !/2 ! 2 !). Thus, the required number = 4 C 2 × 2 C 2 × (6 !/2 ! 2 !)=1080

Q. Number of $6$ digit numbers that can be made with the digits $1, 2, 3$ and $4$ and having exactly two pairs of digits is

978

1801

1080

789

The number will have $2$ pairs and $2$ different digits.
The number of selections $=^{4} C_{2} \times^{2} C_{2}$ ,
and for each selection, number of arrangements
$= \frac{6 !}{2 ! 2 !}$ .
Thus, the required number
$=^{4} C_{2} \times^{2} C_{2} \times \frac{6 !}{2 ! 2 !} = 1080$

Q. Number of 6 digit numbers that can be made with the digits 1,2,3 and 4 and having exactly two pairs of digits is

978

1801

1080

789

The number will have 2 pairs and 2 different digits. The number of selections =4C2​×2C2​, and for each selection, number of arrangements =2!2!6!​. Thus, the required number =4C2​×2C2​×2!2!6!​=1080

Q. Number of $6$ digit numbers that can be made with the digits $1, 2, 3$ and $4$ and having exactly two pairs of digits is

The number will have $2$ pairs and $2$ different digits.
The number of selections $=^{4} C_{2} \times^{2} C_{2}$ ,
and for each selection, number of arrangements
$= \frac{6 !}{2 ! 2 !}$ .
Thus, the required number
$=^{4} C_{2} \times^{2} C_{2} \times \frac{6 !}{2 ! 2 !} = 1080$