Permutations and Combination — Maths STD 11 Science — Question

$\therefore$ Required number of numbers formed $=\frac{6 !}{2 !}$

$=\frac{6 \times 5 \times 4 \times 3 \times 2 !}{2 !}$

= 360 A 6-digit number is to be formed using the same digits that are divisible by 4. For a number to be divisible by 4, the last two digits should be divisible by 4, i.e. 24, 52, 56, 64, 92 or 96. Case I: When the last two digits are 24, 52, 56 or 64. As the digit 9 repeats twice in the remaining four numbers, the number of arrangements =

$\frac{4 !}{2 !}=12$

∴ 6-digit numbers that are divisible by 4 so formed are 12 + 12 + 12 + 12 = 48. Case II: When the last two digits are 92 or 96. As each of the remaining four numbers are distinct, the number of arrangements = 4! = 24 ∴ 6-digit numbers that are divisible by 4 so formed are 24 + 24 = 48. ∴ Required number of numbers framed = 48 + 48 = 96