Question

(a) A game show wants to place 7 distinguishable prizes into 7 distinguishable boxes, with exactly one prize in each box. In how many ways is this possible?

(b) The game show gets rid of 3 boxes, leaving 7 distinguishable prizes and 4 distinguishable boxes. In how many ways can someone place the prizes in the boxes? (Some boxes may be left empty.)

(c) Suppose the boxes are labeled A, B, C, and D, and the rules of the game show stipulate that 1 prize must go in box A, 1 prize must go in box B, 3 prizes must go in box C, and 2 prizes must go in box D. In how many ways can someone place the prizes in the boxes?

Answer 1

• There are 7 distinguishable prizes.
  There are 7 distinguishable boxes.
  The prizes are to put into boxes such that no box remains empty.
  
  Observe that,
  The first prize has 7 choices for box.
  The secnod prize has 6 choices for box.
  The third prize has 5 choices for box.
  ....
  ....
  ....
  The sixth prize has 2 choices for box.
  The seventh prize has 1 choice for box.
  
  Hence, all possible ways these 7 prizes can be placed in 7 different box = 7*6*5*4*3*2*1 = 5040.
  
  
  
• At present,
  There are 7 distinguishable prizes.
  There are 4 distinguishable boxes.
  The prizes are to put into boxes such that boxes may remains empty.
  
  Observe that,
  The first prize has 4 choices for box.
  The secnod prize has 4 choices for box.
  The third prize has 4 choices for box.
  ....
  ....
  ....
  The sixth prize has 4 choices for box.
  The seventh prize has 4 choice for box.
  
  Hence, all possible ways these 7 prizes can be placed in 4 boxes such that some may remain empty = 47 = 16384.
  
  
• Box A has 1 prize.
  Box B has 1 prize.
  Box C has 3 prizes.
  Box D has 2 prizes.
  
  The prize for the Box A can be chosen in 7 ways.
  There are 6 prizes remaining.
  The prize for the Box B can be chosen in 6 ways.
  There are 5 prizes remaining.
  The prize for box C can be chosen in ⁵C₃ ways = 10 ways.
  There are 2 prizes remaining.
  The prize for box C can be chosen in ²C₂ ways = 1 way.
  
  Hence, the total number of ways in which they can be placed = 7*6*10*1 = 420 ways.
  
  

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