Q. HOW MANY 4 DIGIT NUMBER CAN BE FORMED SUCH THAT IT CONTAINS EXACTLY 3 DIFFERENT DIGITS ??

• Case I:
  The repeated digit is the unit's digit.
  So, the 1st, 2nd and 3rd digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 4th digit (unit's digit) can be either equal to the 1st, 2nd or 3rd digit.
  Thus, in all we have:
  9x9x8x3
  
  Case II:
  The repeated digit is the ten's digit.
  So, the 1st, 2nd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 3rd digit (ten's digit) can be either equal to the 1st or 2nd digit.
  Thus, in all we have:
  9x9x2x8
  
  Case III:
  The repeated digit is the hundred's digit.
  So, the 1st, 3rd and 4th digits can be selected in 9 x 9 x 8 ways, respectively.
  Now the 2nd digit (hundred's digit) is equal to the 1st digit.
  Thus, in all we have:
  9x1x9x8
  
  In totality, we have 9x9x8(3+2+1) = 9x9x8x6 = 3888
  
  Hope this helps.
• 10 years agoHelpfull: Yes(34) No(11)
• 4 DIGIT NUMBER WITH EXACTLY 3 DIFFERENT DIGITS i.e repition not allowed
  total no. of digits =10(0,1,2,3,...,9)
  thousand place cant be 0 so can be filled in 9 ways
  rest three places can be filled in 9P3 ways
  total = 9* 9P3 = 9*9*8*7= 4536
  
  
  
  
• 10 years agoHelpfull: Yes(18) No(11)
• answer is 1944 (9.9.8.3=1944)
  1st digit can be selected in 9 ways
  2nd in 9 ways again
  3rd in 8 ways. This way we have 3 distinct digits.
  now the 4th digit has to be a repeat of 1st digit or 2nd or 3rd digit
  Hence it can be done in 3 ways
  So, 9.9.8.3=1944
• 10 years agoHelpfull: Yes(17) No(2)
• ans is 3888
  no of arrangement=3!=6
  one digit should be repeated,n no of possible arrangement is 6
  so 9*9*8*6=3888
• 10 years agoHelpfull: Yes(4) No(0)
• asking for 3 difffer digit
  its can be formed by 2 same & 2 differ digits
  11,22,...,99 => 9(4!/2! *9c2 - 3!/2! *8)
  00 => 9c1*8c1*(3!/2!)
  total=9(4!/2! *9c2 - 3!/2! *8)+9c1*8c1*(3!/2!)
  
  is it right??
• 10 years agoHelpfull: Yes(3) No(0)
• @K Arun Ya u r right.3888 is the right answer.bcoz it is asking for exactly 3 diff digits (not exactly 4 diff digits):
  So answer will be 8424(from my previous calculation)-4536(nos with exactly 4 diff digits)=3888
• 10 years agoHelpfull: Yes(1) No(0)
• @Rakesh: Me too thought that only... but 4536 is not in the option..
  
• 10 years agoHelpfull: Yes(0) No(0)
• @rakesh 1134 is also a no with 3 diff digits but 1 repeated
• 10 years agoHelpfull: Yes(0) No(0)
• 1.Four digit nos with no digits repeated:9x9x8x7=4536
  2.Nos with two zeros(ex:1004): (9C1x8C1x3!)/2!=9x8x3=216
  3.Nos with two 1's:(ex:1123,2311,as these have exactly three different digits 1,2 and 3)
  i.nos starting with 1(ex:1123):9C2x3!=216
  ii.nos not starting with 1(ex:2311):(8C1x8C1X3!)/2!=192
  So total nos with two 1's=216+192=408
  Similarly for two 2's=408,two 3's=408...
  so total nos with exactly three different digits=4536+216+9x408=8424
  @K Arun Is this option present there?
• 10 years agoHelpfull: Yes(0) No(3)
• @satyaranjan: No,
  options are
  3280
  4216
  3888
  Ans might be 3888 not sure
• 10 years agoHelpfull: Yes(0) No(1)
• @k arun could u plz upload more ques?any ques repeated from here?
  
• 10 years agoHelpfull: Yes(0) No(0)
• @K.ARUN :it is impossible because there will be 4536 no if it contains exactly 4 different digits.
• 10 years agoHelpfull: Yes(0) No(0)
• 9*9*8*3=1944? is there any option like this?
• 10 years agoHelpfull: Yes(0) No(0)