How many 6 digit telephone numbers can be formed if each number starts with 35 and no digit appears more than once?

Option
1) 720
2) 360
3) 1420
4) 1680

• 6 digit number where first 2 digit are fixed .
  1*1*8*7*6*5 =1680
• 9 years agoHelpfull: Yes(33) No(0)
• here first two digits are fixes the rest 4 digits have to be selected from 8 digits so 8p4 = 8!/(8-4)!=1680
• 9 years agoHelpfull: Yes(9) No(0)
• out of 6 digits 2 digits are fixed then 4 place are there and we have remaining letters are 1,2,4,6,7,8,9,0
  therfore the total digits are 8 then 8p4=8!/[8-4]!=8!/4!=1680
• 9 years agoHelpfull: Yes(9) No(0)
• ans -1680
  first n second place will be 3 n 5 den next 4 digits will be frm 0-7
  thus 1*1*8*7*6*5=1680
• 9 years agoHelpfull: Yes(2) No(0)
• 35----
  1*1*8*7*6*5=1680
• 9 years agoHelpfull: Yes(2) No(0)
• 8*7*6*5=1680
• 9 years agoHelpfull: Yes(2) No(0)
• as 3 and 5 are fixed so ans wilb for 3 rd place 8 ways , for 4 place 7 ways for 5th place 6 ways and for 6th place 5 ways so 8*7*6*5=1680
• 9 years agoHelpfull: Yes(1) No(1)
• Answer is 360. As it is a permutation problem,the order matters,
  Hence, n be 6 and r be the possibilities.
  nPr = (n!)/(n-r)! =6!/(6-4)! =360
• 9 years agoHelpfull: Yes(0) No(8)
• 6 digit number in which 3 and 5 are fixed
  35----
  so we have only 4 places to fill
  but no digit repeated so the 3rd place will be filled by 8 ways becoz 3 and 5 are filled already
  4th is filled by 7 ways and so on
  35----
  35 8*7*6*5
  1680 ans
• 8 years agoHelpfull: Yes(0) No(0)
• - - - - - -
  6 5 4!=720
• 8 years agoHelpfull: Yes(0) No(4)
• 3*5*6*7*8*9=1680
• 6 years agoHelpfull: Yes(0) No(0)