How many four digit can be formed using 2,3,4,5,6,7 only once divisible by 25
a 12
b 20
c 24
d 40

• Ans : 24
  
  For any number to be divided by 25, it should be ended with {25,20,75,00)...
  Here 0 is not in the list... so d possibilities are the numbers ended with (25 nd 75)
  
  if v take the no. to end with 25, the remaining 2 digits can be any 2 among {3,4,6,7} . so the possibities are 4!/2= 24/2= 12
  
  if v take the no. to end with 75, the remaining 2 digits can be any 2 among {2,3,4,6} . so the possibities are 4!/2= 24/2= 12
  
  so 12+12=24
  
  
• 10 years agoHelpfull: Yes(61) No(9)
• ans is 20 as total 24 case - (3125,4375,5625,one more )
• 10 years agoHelpfull: Yes(5) No(1)
• ans 20.
  
  For any number to be divided by 25, it should be ended with {25,50,75,00)
  Here 0 is not in the list. so d possibilities are the numbers ended with (25 nd 75)
  
  if v take the no. to end with 25, the remaining 2 digits can be any 2 among {3,4,6,7} . so the possibities are 4!/2= 24/2= 12
  
  if v take the no. to end with 75, the remaining 2 digits can be any 2 among {2,3,4,6} . so the possibities are 4!/2= 24/2= 12
  
  so 12+12=24
  
  but 4 cases out of these 24 are such as being divided by 25 twice
  these nums are 3125, 4375, 5625, 6875.
  
  so excluding these fours , we get 24-4=20.
  
• 10 years agoHelpfull: Yes(5) No(0)
• 12 is d answer
• 10 years agoHelpfull: Yes(4) No(10)
• 24 ans nahi h because 4375 divisible 25 in 2times
• 10 years agoHelpfull: Yes(4) No(4)
• according to my logic ans is 12 because according to question 2,3,4,5,6,7 is to be used once ( repetition) not allowed.
  note- a no. to be divisible by 25 should be ended with 25,50,75,100 but however in question we are not having zero,therefore we are restricted to 25 and 75 only.
  thus,
  4 positions can be filled by
  3*2*2*1 = 12
  proper explanation-
  digits in
  4th digit position = 3 or 4 or 6 (i.e 3 ways)
  3rd digit position = 4 or 6( if 3 is taken to be in the 4 th place) (i.e 2 ways)
  2nd digit position = 2 or 7 (i.e 2 ways)
  1st digit position = 5 ( i.e 1 way)
  
  therefore ans is 3*2*2*1 = 12
  
• 10 years agoHelpfull: Yes(1) No(5)
• anuj your way of solving is right but you have forgotten 1 number 6875 so 24 - (3125, 4375, 5625, 6875)
  so ans is 20
• 10 years agoHelpfull: Yes(0) No(0)
• @arpit jain 6875 is not that no as 8 is not provided to us . So ,the solution you provided is wrong .
• 10 years agoHelpfull: Yes(0) No(0)
• 5 multiples are 25
  50---0 is not given in question so eliminated 50.
  75
  100---0 not given 100 eliminated.
  for 5 to be the last digit ,,last but one must be either 2 or 7.
  for --25
  4p2=12
  for --75
  4p2=12
  so 24.
• 10 years agoHelpfull: Yes(0) No(1)
• what about repetition of numbers?
• 10 years agoHelpfull: Yes(0) No(1)
• Ans : 24
  
  For any number to be divided by 25, it should be ended with {25,50,75,100)...
  Here 0 is not in the list... so d possibilities are the numbers ended with (25 nd 75)
  
  if v take the no. to end with 25, the remaining 2 digits can be any 2 among {3,4,6,7} . so the possibities are 4!/2= 24/2= 12
  
  if v take the no. to end with 75, the remaining 2 digits can be any 2 among {2,3,4,6} . so the possibities are 4!/2= 24/2= 12
  
  so 12+12=24
  
• 10 years agoHelpfull: Yes(0) No(3)
• as no should be devisible by 25 so the last two position must be 25 or 75.
  when we consider _ _ 2 5 then for first two places.
  select two numbers out of 4 i.e 4c2*2!(2! for arranging these).
  when _ _ 7 5 then select 2 out of 4 i.e 4c2*2!(for arranging the two selected numbers).
  so total ways= 4c2*2! + 4c2*2!
  =6+6
  =12
• 10 years agoHelpfull: Yes(0) No(2)