find the terms which do not contain 7 between 1000 and 9999

• Total number of four digit numbers =9000 (i.e 1000 to 9999 )
  We try to find the number of numbers not having digit 2 in them.
  Now consider the units place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
  Tens place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
  Hundreds place it can be selected in 9 ways (i.e 0,1,2,3,4,5,6,8,9)
  Thousands place can be selected in 8 ways (i.e 1,2,3,4,5,6,,8,9) here '0' cannot be taken
  Total number of numbers not having digit 7 in it =9 x 9 x 9 x 8 =5832
  Also,Total number of numbers having digit 7 in it = 9000-5832 =3168
  
• 8 years agoHelpfull: Yes(18) No(1)
• could you pls expain how 8*9*9*9 ?
• 9 years agoHelpfull: Yes(10) No(2)
• total no of terms do not contain 7=8*9*9*9=5832
• 9 years agoHelpfull: Yes(8) No(2)
• this question is of permutation and combination
  [] x[]x[]x[]
  8 9 9 9
  thererfore 8x9x9x9=5832
• 9 years agoHelpfull: Yes(7) No(2)
• from 1000 to 9000 there are the part of 7000-7999 is eliminated so 1000-9000 without 7 is 8 terms.
  now from 100-999 the part of 700-799 will be eliminated so no. of terms without 7 is 9.
  similarly between 10 to 99 it is also 9 and 1-10 it is also 9. so 8*9*9*9 is the ans
• 9 years agoHelpfull: Yes(7) No(3)
• Total no of no.=9000
  total no of term without 7=8×9×9×9=5832
  Ans=9000-5832=3168
• 9 years agoHelpfull: Yes(6) No(21)
• y do you finally subtract 5832 from 9000..pls explain any one
  
• 9 years agoHelpfull: Yes(4) No(2)
• Draw four boxes in a piece of paper
  The 4th box can be filled with 1,2,3,4,5,6,8,9=8 ways (as 0 and 7 can't be included in thousand place)
  the 3rd,2nd,1st boxes can be filled by 9 ways each(0,1,2,3,4,5,6,8,9)
  So the total no of ways=9*8*8*8=5832
• 8 years agoHelpfull: Yes(3) No(0)
• thanku@brajeshrai
• 9 years agoHelpfull: Yes(2) No(3)
• detailed explanation plzz.
• 9 years agoHelpfull: Yes(2) No(0)
• Total no of numbers=9000
  Total no of numbers with out 7=8c1*9c1*9c1*9c1=5832
  no of terms =9000-5832=3168
• 9 years agoHelpfull: Yes(2) No(1)
• 8_8_7_6
  8*8*7*6=2688
• 8 years agoHelpfull: Yes(0) No(1)
• units place it can be any number except 7 -->9C1
  same with tens and hundreds place.
  1st digit cannot be 0 or 7 .So 8C1
  total= 8C1 *9C1 *9C1 *9C1=8*9*9*9
• 8 years agoHelpfull: Yes(0) No(0)
• the numbers whis are contain 7 is for unit place 1000 ways and for ten place 1000 and for hundred place 1000 and for thuusend place 999 so total 3999 .total no. present in between 1000 to 9999 is 9000 so the terms whis are not containing 7 is 9000-3999=5001
• 8 years agoHelpfull: Yes(0) No(0)
• why we have subtract it from 9000?
  
• 8 years agoHelpfull: Yes(0) No(0)