in how many we can arrange 7 different things to 3 people.all of them must get at least one

• (7-1)C(3-1)
  6C2
• 11 years agoHelpfull: Yes(21) No(6)
• n-1cr-1 fr identical objects
  6c2=15
• 11 years agoHelpfull: Yes(8) No(1)
• Formula for atleast one is= n-1Cr-1;
  so the answer is 6C2= 15;
• 11 years agoHelpfull: Yes(6) No(1)
• we can use permutation formula for arranging this..so 7p3=210
• 11 years agoHelpfull: Yes(3) No(7)
• 1806
  here thigs are different so we cannot use n-1c(r-1)
  the different way of grouping 7 different things are-(1/3/3),(1/4/2),(1,5,1),(2,2,3)
  (1/2/3)=(7!*3!)/(3!*3!*2!)=30240/72=420
  (1/4/2)=(7!*3!)/(4!*2!)=30240/48=630
  (1/5/1)=(7!*3!)/(5!*2!)=30240/240=126
  (2/2/3)=(7!*3!)/(2!*2!*2!*3!)=30240/48=630
  on addind all the values we get 1806
• 11 years agoHelpfull: Yes(3) No(7)
• arranging 7 different things to 3 people= first select and then rearrenge as:
  3^7 * 3!
• 11 years agoHelpfull: Yes(2) No(5)
• If it is given there are different objects n to be distributed among different persons r . The formula is r^n. Here r is 3 and n is 7 so 3^7.
  But as given each one must get at least one so we have (0 _ _ ) * 3! cases and (0 0 _)*2! to eliminate
  So ans is 3^7- 3!*2!= 2187-12=2175
• 11 years agoHelpfull: Yes(1) No(0)
• n-1Cr-1 is for similar things;repetation allowed it is 7*7*7 as each one of then can get either(1,2,3,4,5,6,7).
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 299
  first factorize 725 u get 29*5*5*1*1
  i.e. 39+63+63+67+67=299
• 11 years agoHelpfull: Yes(0) No(5)
• first allote 1 to each that can be done in 3! ways
  the distribute 4 thing in 3 4p3
  4p3*3!
  
  
• 11 years agoHelpfull: Yes(0) No(3)
• ans is 2646.
  
• 11 years agoHelpfull: Yes(0) No(2)
• 7*7*7 is the answer.. 3 place can be arranged in 7 ways each.. would it had been at most 1.. den d nswr ws supos to be 7p3 bt right now its 7*7*7!!
  since its a question of arrangemnt combination is not supposed to be use here!
• 11 years agoHelpfull: Yes(0) No(0)
• Answer is 1491.....that is 1/2/4, 1/3/3 ,1/1/5, 2/2/3 gives combinations of 3*105+3*140+3*42*3*210=1491
• 11 years agoHelpfull: Yes(0) No(0)