How many positive numbers not greater than 4300 can be formed using the digits 0,1,2,3,4 where repeatation is allowed?

Options: a)576 b)575 c)620 d)640

• a) 576
  
  1 digit = 5
  2 digits = 4* 5 = 20
  3 digits = 4*5*5= 100
  
  Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
  0,1,2,3,4 in hundreds ,tens and ones place
  I.e 3*5*5*5 = 375
  
  Now From 4000 to 4299 = 1*3*5*5 =75
  
  
  Total = 5+20+100+375+75+1 =576 (1 for 4300)
  
• 9 years agoHelpfull: Yes(41) No(10)
• When repitition is allowed
  
  1 digit = 5
  2 digits = 4* 5 = 20
  3 digits = 4*5*5= 100
  
  Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
  0,1,2,3,4 in hundreds ,tens and ones place
  I.e 3*5*5*5 = 375
  
  Now From 4000 to 4300 = 1*3*5*5 =75
  
  
  Total = 5+20+100+375+75 =575
• 9 years agoHelpfull: Yes(8) No(4)
• 4300
  1,step o*5^0=0
  2,step 0*5^1=0
  3,step 3*5^2=75
  4,step 4*5^3=500
  some 0+0+75+500=575.
  ans 575
• 9 years agoHelpfull: Yes(8) No(0)
• 0 is not a positive number.. the answer would be 575
• 9 years agoHelpfull: Yes(5) No(0)
• Answer is 575
• 9 years agoHelpfull: Yes(2) No(5)
• (4c1 * 5c1 * 5c1 * 5c1) + (1c1 * 3c1 * 5c1 * 5c1 ) + (1c1 *1c1 *1c1 *1c1 )= 500+75+1=576
• 9 years agoHelpfull: Yes(2) No(3)
• b) 575
  
  1 digit = 5
  2 digits = 4* 5 = 20
  3 digits = 4*5*5= 100
  
  Now,4 digit number till 4000 can have digits 1,2,3 in thousands place ( total choices = 3 )
  0,1,2,3,4 in hundreds ,tens and ones place
  I.e 3*5*5*5 = 375
  
  Now From 4000 to 4300 = 1*3*5*5 =75
  
  
  Total = 5+20+100+375+75 =575
• 9 years agoHelpfull: Yes(1) No(4)
• first count for less than 3400.....we get 4x5x5x5 =500
  
  now from 4000 to 4299....we get 1x3x5x5 = 75
  
  now 4300 is left which we can write in 1 way =1
  
  adding above 500+75+1 =576
• 9 years agoHelpfull: Yes(1) No(2)
• kaun best solution ka muhar maarta hai bhai kch dekh lia kar
• 9 years agoHelpfull: Yes(1) No(2)
• One easy method is to notice that there is a one-to-one correspondence between the numbers less than 4300 that can be made from the digits 0–4 and all the numbers less than 43005 in their base-5 representation. The answer is therefore 43005=4×5^3+3×5^2=575.
• 9 years agoHelpfull: Yes(0) No(0)
• 4300=4×5^3+3×5^2=575.
• 7 years agoHelpfull: Yes(0) No(0)