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(#M40019252) GMATE question PermN Comb Keep an EYE

Q. How many four-digit numbers,that are divisible by 4 can be formed, using the digits 0 to 7 if no digit is to occur more than once in each number

Asked In Gmate Ish Mishra (12 years ago)

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sushma  (12 years ago)

470
last 2digts must be 04,12,16,20,24,32,36,40,52,56,60,64,72,76
with 0's in last two digits:(04,20,40,60) - - --
6*5*4 =120
0 in middle: - - --
5*1*10=50
without any0:- - --
5*4*10=200
total 470

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deepak  (12 years ago)

there are 8 no.s possible.
because the no. is divisible only if last two digits are divisible by 4.
i.e __04,
__12,
__16,
__20,
__24,
__32,
__36,
__40.

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Ajithkumar  (12 years ago)

for a no. to be divisible by 4, last two digits must be divisible by 24
so last 2digts must be 04,12,16,20,24,32,36,40,52,56,60,64,72,76.
total = 14
in a 4 digit no. to end with 04
the first digit has these many choices
1,2,3,5,6,7 = 6 choices
then obviously second digit has 6 choices
3 & 4 digits must be 0 & 4 which is equal to 1 choice.
therefore, for 04 ending nos., the chances are 6*6*1*1 =36.
den for 14 types of ending 2 digit nos., there are 14*36=504 possibilities.

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