Capgemini
Company
Numerical Ability
Permutation and Combination
How many 5 digit no. can b formed wit digits 1, 2, 3,4,5,6 which r divisible by 4 and digits not
repeated
Read Solution (Total 9)
-
- Number can bedivisible by 4 if its last 2 digits are divisible by 4 so the possiblities are
12,16,24,32,36,52,56,64= 8 possiblities
Since 2 digits are occupied remainin g 3 blanks can be filled by 4 remaining digits i.e 4!
Therefore the ans is 4!*8
24*8=192 - 10 years agoHelpfull: Yes(18) No(0)
- number can divisible by 4 is 8 ways (12,16,24,32,36,52,56,64)
last 2 digits occupied by 8 ways
remaining 3 letters can be possible to filled in 4p3 ways
i.e 4p3 * 8 = 24*8 = 192
- 10 years agoHelpfull: Yes(2) No(1)
- Vamshi priya----digits should be not repeated that is mentioned in question...so 192 is right answer.
- 9 years agoHelpfull: Yes(2) No(0)
- The question is for without repetition so we shouldn't take 44 into account
- 9 years agoHelpfull: Yes(2) No(0)
- Number can bedivisible by 4 if its last 2 digits are divisible by 4 so the posiblities are
12,16,24,32,36,44,52,56,64= 9 posiblities
Since 2 digits are occupied remaining 3 blanks can be filled by 4 remaining digits i.e 4!
Therefore the ans is 4!*9
24*9=216
- 10 years agoHelpfull: Yes(1) No(13)
- Why have u taken 4! @All
- 8 years agoHelpfull: Yes(1) No(0)
- pls explain
- 10 years agoHelpfull: Yes(0) No(1)
- @VAMSHI PRIYA -
NOTE : NUMBERS ARE NOT REPEATED SO, 44 WILL NOT BE TAKEN INTO CONSIDERATION !!
4!*8 =192 - 8 years agoHelpfull: Yes(0) No(0)
- - - - - -
( last 2 digits 1's and 10's place divisible by 4 i.e 12,16,24,32,36,52,56,64 total 8 ways(without repetition))
( 3rd digit 100th place 4 ways because already 2 digits are used)
(4th place 1000's place 3 ways )
(5th digit 10000's place 2 ways)
so ans is 8ways*4ways*3ways*2ways =192 - 8 years agoHelpfull: Yes(0) No(0)
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