A 6-digit number is to be formed using digits of 659942, in which 9 repeats twice.
∴ Required number of numbers formed = \(\frac{6!}{2!}\) = \(\frac{6\times5\times4\times3\times2!}{2!}\) = 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
