hw many 6 dgts numbrs cn be made using digits 1,2,3,4,5,6 without repetiotin such that hundread's digit is greater than ten's digit and ten's is greater than ones digit??
Give any short hand explanation if possible.

• let hunderds place with 3 then possible cases
  3: 3 2 1
  4: 4 3 2
  4 3 1
  4 2 1
  5: 5 4 3
  5 4 2
  5 4 1
  5 3 2
  5 3 1
  5 2 1
  6: 6 5 4
  6 5 3
  6 5 2
  6 5 1
  6 4 3
  6 4 2
  6 4 1
  6 3 2
  6 3 1
  6 2 1
  so total combinations : 20*6=120
  
• 10 years agoHelpfull: Yes(41) No(3)
• 6 numbers can be arranged in 6! ways and according to the condition Hundereds>tens>ones they cannot interchange there postions so they are identical an can be arranged in 3! ways so,
  6!/3!=120
• 10 years agoHelpfull: Yes(16) No(1)
• the correct answer is 180.
  last 2 digit 21=24
  last 2 digit 31=18
  last 2 digit 41=12
  last 3 digit 651=6
  last 2 digit 32=18
  last 2 digit 42=12
  last 2 digit 652=6
  last 2 digit 43=12
  last 3 digit 653=6
  last 3 digit 654=6
  total-120
  
• 10 years agoHelpfull: Yes(3) No(2)
• there are 6 numbers 1,2,3,4,5,6
  
  leaving tens hundreds and units place the remaining three places could be filled in
  
  6*5*4*(hundreds>tens>unit)
  
  so the last three places could be filled in only 1 way if 3 digits are left
  
  so the answer is 6*5*4=120
• 10 years agoHelpfull: Yes(3) No(0)
• step1 :-you have 6 place last 3 digit have some condition
  but starting 3 digit can be arrange by 3*2*1=6
  step2 :-at hundred position only 4 option available=6,5,4,3
  (we can not include 2 and 1 because of condition ,give in question)
  step3 :-if we place 6 at hundred position then at ten's digit we can put only
  one of the 4 number and at unit digit we have 4 option too
  so total 1*4*4=16 posibility.
  
  4 case and posible combinition...
  100 10 unit
  6(number) 4 4 =1*4*4=16
  5 3 3 =1*3*3=9
  4 2(3 or 2) 2(2 or 1)=1*2*2=4
  3 1(only 2) 1(only 1)=1*1*1=1
  
  total=30 combination for last 3 digit and we know 6 combination for stating
  digit.
  so
  
  ans is =6*30=[180] ...:
• 10 years agoHelpfull: Yes(2) No(13)
• what is the correct answer?
• 10 years agoHelpfull: Yes(2) No(0)
• CORRECT ANSWER IS 120
• 10 years agoHelpfull: Yes(2) No(2)
• we have to select 6 number in which condition is to be satisfied first so last three digit can be slected in 6c3 ways and remaining 3 digit can be occupied in 3 places with 3! so simultaneously it can be placed in 3!*6c3 ie 120 is the ans
  
• 10 years agoHelpfull: Yes(2) No(0)
• sorry for my previous ans.....
  In one's digit ----> 1,2,3,4 = 4 numbers can b placed
  In ten's digit ----> 2,3,4,5 = 4 numbers can b placed
  In hundreads's digit ----> 3,4,5,6 = 4 numbers can b placed
  
  In rest 3 places 1,2,3,4,5,6 shpuld b placed in 6c3=20 ways
  So, total 6 dgts number cn b made=4+4+4+20=32 ways...(ans.)
  
• 10 years agoHelpfull: Yes(1) No(2)
• N articles (n1,n2,n3 .... nN) to be arranged in N places such that ni comes before nj, nj comes before nk then...
  
  Total no of articles have restricted to their positions is = 3 -->>(ni,nj,nk)
  
  
  then no off possibilities is ==>>> (n C 3)*(n-3)!
  
  6 C 3* 3! === 120
• 10 years agoHelpfull: Yes(1) No(1)
• Ans is 22*6:132
• 10 years agoHelpfull: Yes(0) No(8)
• In one's digit ----> 1,2,3,4 = 4 numbers can b placed
  In ten's digit ----> 2,3,4,5 = 4 numbers can b placed
  In hundreads's digit ----> 3,4,5,6 = 4 numbers can b placed
  
  In rest places 1,2,3 shouldn't b placed....
  So 6,5,4 should b placed in rest places in 3!=6 ways
  
  So, total 6 dgts number cn b made=4+4+4+6=18 ways...(ans.)
  
  
  
  
• 10 years agoHelpfull: Yes(0) No(4)