Latest Capgemini Aptitude Question SOLUTION: A 5-digit number is formed by the digits 2,4,5,6,8 (each digit used exactly once) . What is the probability that the number formed is divisible by 4 ?

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A 5-digit number is formed by the digits 2,4,5,6,8 (each digit used exactly once) . What is the probability that the number formed is divisible by 4 ?

Asked In Capgemini jayakishore (10 years ago)

Solved RAKESH Read Solution (14)

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Read Solution (14) :

RAKESH  (10 years ago)

a no. is divisible by 4 if last two digit is divisible by 4
so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
last two places are fixed by 24,28,48,52,56,64,68,84
so remaining 3 places can be filled in 3! ways for each
total=8*3!=48=n(E)
n(S)=5C4=120
p=48/120=2/5

Is this solution Helpfull? Yes (46) | No (3)  

ganesh chemanchula  (10 years ago)

numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
remaining 3 places can be filled in 3! ways....total=8*9=48
probability=(48/120)=2/5

Is this solution Helpfull? Yes (17) | No (3)  

Nitesh Dilip Kadam  (9 years ago)

Arranging 4 or 8 at the end of the required digit to make it divisible by 4....2C1

Remaining 4 places are to be filled by four numbers....4P4(as order is important)
req answer=2C1*4P4/5P5 (5P5 is sample space)
=2*4!/5!
=48/120
=2/5

Is this solution Helpfull? Yes (5) | No   

kunda anusha  (9 years ago)

here all the below process is correct.but 5c4=5 but not 120

Is this solution Helpfull? Yes (4) | No   

Rakesh Chepuri  (9 years ago)

actually all are written 5C4........................Its not 5C4 ...it is 5P4......

in this problem we are arranging the digits not selecting.

and 5C4 value is .........5 not 120

Is this solution Helpfull? Yes (4) | No   

Mohammad Umar Khan  (9 years ago)

a no. is divisible by 4 if last two digit is divisible by 4
so numbers ending with 24,28,48,52,56,64,68,84 are divisible by 4
last two places are fixed by 24,28,48,52,56,64,68,84
so remaining 3 places can be filled in 3! ways for each
total=8*3!=48=n(E)
n(S) = 5!
so
asn= n(E)/n(S)
=48/120=2/5

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Chirag Ranjan Lahiri  (9 years ago)

total=8*3!=48=n(E)
n(S)=5C4=120
p=48/120=2/5

Is this solution Helpfull? Yes (2) | No   

adityesh mohanty  (10 years ago)

LAST 2 nos must be divisible by 4.
24,28,48,52,56,64,68,84
and rest 3 places can be filled = 3*2*1 ways(as no repetition)
so total = 8*6=48
5C4=120
p=48/120.

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MONIKA   (9 years ago)

for the no. should be divisible by 4 the cases are 24,28,48,64,68 and 84 so no. of cases possible be 6X3! and total no. of cases be 5! so p=3/10

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pooja singh  (9 years ago)

120 ways..

Is this solution Helpfull? Yes (1) | No (1)  

Devendra Marghade  (10 years ago)

1/4
Numbers ending with 24,28,48,84,56 will be divisible by 4.
Total numbers ending with above last 2 digits combination=(3!)*5=30
Total 5 digit Numbers formed with 5 digits without repetition=5!=120
So probability=30/120=1/4

Is this solution Helpfull? Yes | No (11)  

alekhya.rangannagari  (9 years ago)

25648
divisible by 4

Is this solution Helpfull? Yes | No (1)  

Rashmi Ranjan  (9 years ago)

ans is : 48

Is this solution Helpfull? Yes | No (1)  

Vikash kumar  (7 years ago)

correct answer=48
the condition of divisible of 4=the two unit digit divisible by 4
so total no=24/28/48,52,56,64,68,84 so total=8
accourding to question repeatation is not allow so
3p3*8=48

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