Find no of ways in which 4 particular persons a,b,c,d and 6 more persons can stand in a queue so that A always stand before B. B always stand before C, And C always stand before D.

• a,b,c,d are grouped ie consider them as one and remaining as 6.
  total 6+1=7! ways
• 10 years agoHelpfull: Yes(36) No(18)
• 10C4*6!
  first choose 4 plces out of 10 and place ABCD as mentioned which can't be arranged further(i.e they have fixed pattern) so now we have only 6 persons to arrange so 6!.
• 10 years agoHelpfull: Yes(24) No(2)
• it is not given that A will stand just before B so _ A _ _ B _ C _ D etc sequences can be happen.
  so for this choose any 6 locations and place the 6 other persons(i.e 10C6) then at remaining 4 position A,B,C,D will arranged automatically according to the condition.so ans is 10c6*6!
  
• 10 years agoHelpfull: Yes(7) No(2)
• Take abcd as one unit.And other 6 persons.
  So we have total 7 units.
  To arrange them in different ways:::No of ways=7!
• 10 years agoHelpfull: Yes(5) No(3)
• no of ways=7!*1
• 10 years agoHelpfull: Yes(2) No(2)
• Ans will be 7!.
  Reason: The arrangement A B C D is fixed and there are 6 more persons. Taking A B C D as one, there are 7 persons.
  7 persons can stand together in 7! ways and therefore answer is 7!
• 10 years agoHelpfull: Yes(2) No(2)
• (abcd)!+6! equal to 7!.ans is 7!
  
  
• 10 years agoHelpfull: Yes(2) No(0)
• 10p6 because abcd not give arrange togather
  
• 10 years agoHelpfull: Yes(2) No(2)
• ans is 10c4+7!
  chances of ABCD I among 10 persons is 10c4.. and 7 person(abcd together considerd as 1) is 7!..we add it .. not mul that.
  if more confuse consider ABC and just 2 more person (X,Y) solve same method.A STAND BEFORE B,B STD BF C.
  ans is 5C3+3!=16.
  16 possible ways are
  XYABC,XAYBC,XABYC,XABCY
  YXABC,YAXBC,YABXC,YABCX
  AXYBC.AXBYC,AXBCY
  AYXBC,AYBXC,AYBCX
  ABCXY,ABCYX.
• 10 years agoHelpfull: Yes(2) No(1)
• 6!(for arranging 6 pepole)
  7 option to place abcd taking 1 unit in between 6 person.
  so ans=7*6!
• 10 years agoHelpfull: Yes(1) No(7)
• 7! *2 is the ans bcoz they can be arranged in 2 ways /
• 10 years agoHelpfull: Yes(1) No(5)
• totally ten people are there.....combinations of places are...(1234),(2345),(3456),(4567),(5678),(6789),(78910).....so 7 is the answer
• 10 years agoHelpfull: Yes(1) No(2)
• (d+c+b+a)=1
  remaining 6 persons can stand in 6!ways
  so 1+6=7!
• 10 years agoHelpfull: Yes(1) No(0)
• 10C4*7!
  first choose 4 plces out of 10 and place ABCD as mentioned which can't be arranged further(i.e have fixed pattern) so now we have 7 peoples to arrange so 7!.
• 10 years agoHelpfull: Yes(1) No(0)
• We have to arrange only left 6 more person.. 10p6= 10!/4!
• 9 years agoHelpfull: Yes(1) No(0)
• can u explain it
• 10 years agoHelpfull: Yes(0) No(0)
• Manish sr what is the correct answer for this??
  
• 8 years agoHelpfull: Yes(0) No(0)
• For a,b.c.d=1C1
  Therefore=1C1*1C1*1C1*1C1*6!
• 7 years agoHelpfull: Yes(0) No(0)
• Here there are 10 persons in total. Out of which the position of 4 is always fixed(A->B->C->D). Therefore there is no chance of any other arrangements for them. Therefore we can take them as one unit. Now out of 10, after grouping the total remains 10-3units(as the 4 unit has now become only one unit, therefore the number of unit considered is only 1). Therefore it's 7 remaining,. Now as we know the number of ways in which we can arrange 7 unit is 7!, i.e on the first position,any of the seven unit, next any of remaining 6, third any of remaining 5, 4th any of remaining 4, 3rd any of remaining 3,2nd any of remaining 2 and at last the only left person. This is same as 7*6*5*4*3*2*1=7!
• 5 years agoHelpfull: Yes(0) No(0)
• Here the total persons are 4+6=10
  So there are 10C4*6! ways=151200 ways
• 2 years agoHelpfull: Yes(0) No(0)