Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits 1,3,2,6,8,9 is equal to

Question

Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits 1,3,2,6,8,9 is equal to (A) 192 (B) 32 (C) 1152 (D) 384

Tardigrade Admin · Accepted Answer

A number is divisible by four, if the last two digit are divisible by four. In this case last two digits can be 12, 16, 28, 32, 36, 68, 92 or 96. Total number of such numbers =8( 4 C 3 ⋅ 3 !)=192

Q. Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits ${1, 3, 2, 6, 8, 9}$ , is equal to

192

32

1152

384

A number is divisible by four, if the last two digit are divisible by four. In this case last two digits can be $12, 16, 28, 32, 36, 68, 92$ or $96$ .
Total number of such numbers $= 8 (^{4} C_{3} \cdot 3!) = 192$

Q. Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits {1,3,2,6,8,9}, is equal to

192

32

1152

384

A number is divisible by four, if the last two digit are divisible by four. In this case last two digits can be 12,16,28,32,36,68,92 or 96. Total number of such numbers =8(4C3​⋅3!)=192

Q. Total number of 5 digit numbers having all different digits and divisible by 4 that can be formed using the digits ${1, 3, 2, 6, 8, 9}$ , is equal to

A number is divisible by four, if the last two digit are divisible by four. In this case last two digits can be $12, 16, 28, 32, 36, 68, 92$ or $96$ .
Total number of such numbers $= 8 (^{4} C_{3} \cdot 3!) = 192$