How many six-digit numbers can be made by using the first 6 natural numbers such that the digit at the unit’s place is greater than the digit at the hundred’s place and the number thus formed is a multiple of 4? (Repetition of digits is not allowed.)

• nos will be XXXHTU
  where U > H
  and TU is multiple of 4.
  
  It is possible when
  U is 2 and H=1, T=3.... such numbers are 3*2*1=6
  U=4, H=1, T=2 ........ such numbers are 3*2*1=6
  U=4, H=3, T=2... such numbers are 3*2*1=6
  
  U=6, T=3, H=1,2,4,5 ... 4*3*2*1=24
  
  so total 42 nos.
• 13 years agoHelpfull: Yes(0) No(1)