If 4 toys are randomly distributed among 4 children, what is the

Question

If 4 toys are randomly distributed among 4 children, what is the probability that no child will get more than two toys?

(A) 1/16
 
(B) 13/64
 
(C) 13/16

(D) 51/64
 
(E) 15/16

Source:  Marty Murray Coaching

gmatophobia · Answer

Let's find the opposite of what's asked i.e. the number of ways when a child can get more than two toys.

Case 1 : 3 + 1 + 0 + 0

The four toys are distributed in a manner such that out of four children, one child gets three toys, another child gets one toy and the remaining two children get no toys.
 
 Number of ways of selecting two children out of four children =  ^4C_2 = 6   
 Number of ways of grouping the toys into bundles of 3 toys and 1 toy=  ^4C_3 * ^1C_1 = 4  (i.e. Out of four toys, select three toys. From the remaining one toy, select one) 
 Number of ways of distributing the grouped toys = 2 (i.e. either the first child gets three toys and the second child gets one, or vice versa)

Total =  48

Case 2 : 4 + 0 + 0 + 0

The four toys are distributed in a manner such that out of four children, one child gets all four toys, and the other three children get none.
 
 Number of ways of selecting one child out of four children =  ^4C_1 = 4  
 Number of ways of grouping the toys =  ^4C_4 = 1  (i.e. Out of four toys, select all) 
 Number of ways of distributing the grouped toys = 1 (i.e. the child selected in the first step gets all toys)

Total =  4 * 1 * 1 = 4

Required Probability

The number of ways when a child can get more than two toys =  48 + 4 = 52  
 The number of ways of randomly distributing 4 toys among 4 children =  4^4 = 256  
 The number of ways of randomly distributing 4 toys among 4 children such that no child will get more than two toys = 256 - 52 = 204

Required Probability =  \frac{	ext{Favourable Ways}}{	ext{Total Number of ways}} = \frac{204}{256} = \frac{51}{64}

Option D

DrAnkita91 · Answer

why is the formula (n+r-1 _C_ r-1) not working here?
4 toys are randomly distributed among 4 children = 7 _C_ 3 ways right?
So denominator should be this?
please correct me

gmatophobia · Answer

DrAnkita91  - This formula works when the toys are identical. The question doesn't mention that.

Maxsparrow · Answer

Antithesis:
Scenario when 1 kid get > 2 toys:

I. 1 kid get 3 toys and the other one get the rest

P1 = 4C1kid x 4C3 toy x 3C1 kid x 1C1 toy
 = 4 x 4 x 3 x 1 = 3 x 4^2

II. 1 kid get 4 toys

P2 = 4C1 kid x 4C4 toy
 = 4 x 1 = 4

Total P Ant= 3x4^2 + 4 = 4 (3x4 + 1) = 4 (13)

Total ways of distributing toys = 4 toys to 4 kids = 4^4

Probability of 1 kid get =<2 : 1- Total P Ant / Total Ways

Prob = 1 - {4 x 13 / 4^4)
 = 1 - 13/64
 = 51/64

ANS D

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