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globaldesi

Why have you not considered
_ _ 2 0
_ _ 3 2
_ _ 4 0

In which case there will be another 16 such possibilities and the answer would then be 66 instead

Could you please explain what m I missing

Posted from my mobile device
yeah you are correct
total cases formed will be
-- 32 = 4
--12= 4
--04=6
--20=6
--40=6
--24=4
total 30
hence not divisible be will be 66
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J2S2019
There are 5 Digits (0,1,2,3,4). How many 4 digit numbers formed by those digits(repetation not allowed) are not divisible by 4?

A.96
B.30
C.66
D.82
E.78

Others:- TIME
i think the OA is incorect
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J2S2019
There are 5 Digits (0,1,2,3,4). How many 4 digit numbers formed by those digits(repetation not allowed) are not divisible by 4?

A.96
B.30
C.66
D.82
E.78


CONCEPT:For a Number to be divisible by 4, Number formed by Last two digits (Unit and tens) of the number MUST be divisible by 4

Last two digits of such 4 digit number may be
_ _ 04
_ _ 12
_ _ 20
_ _ 24
_ _ 32

_ _ 40

Blue cases {3 cases 04, 20, 40} are similar cause they have used digit 0 in last two digits
so number of ways to fill Thousands and hundred digits = 3*2 (3 choices for thousands and 2 choices for hundreds digits) = 6 ways
i.e. Total cases = 3*6 = 18

Red cases {3 cases 12, 24, 32} are similar cause they have NOT used digit 0 in last two digits
so number of ways to fill Thousands and hundred digits = 2*2 (2 choices for thousands cause 0 can't be taken there and 2 remaining choices for hundreds digits) = 4 ways
i.e. Total cases = 3*4 = 12

Total Numbers divisible by 4 = 18+12 = 30 cases

Total 4 digit numbers using 5 Digits (0,1,2,3,4) = 4*4*3*2 = 96

Required cases (Not divisible by 4) = 96 - 30 = 66

Answer: Option C

J2S2019 Bunuel The Answer should be Option C and requires a correction

Also, Options should be written in increasing/decreasing order while I see randomness in them
.
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J2S2019
There are 5 Digits (0,1,2,3,4). How many 4 digit numbers formed by those digits(repetation not allowed) are not divisible by 4?

A.96
B.30
C.66
D.82
E.78

Others:- TIME

to determine How many 4 digit numbers formed by those digits(repetation not allowed) are not divisible by 4 = TOTAL CASES- Total cases divisible by 4

for divisibility of 4 ; we need to have last two digits which are twice divisible by 2 ;
in this case those two digits would be
12 , 20, 40,04,24, 32
and total possible cases would be ; 4*4*3*2*1 ; 96
for no divisible by 4 possible combinations
case 1; when we have 20,40,04 ; 3*2*1*1*1 ; 6 *3 ; 18
case 2; when we have 12,24,32 ; 2*2*1*1* 1; 4*3 ; 12
total cases ; 18+12; 30
so digit numbers not divisible by 4 ; 96-30 ; 66
IMO C

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