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Since no one answered my last question I'm going to post it again. 

I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different? Same means indistinguishable and different means distinguishable.

Thanks!

MIRB16  Mar 6, 2018
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3+0 Answers

 #1
avatar+19344 
+4

Since no one answered my last question I'm going to post it again. 

I have 3 pieces of candy to place in 4 lunch boxes.

In how many ways can I do this if exactly two of the candies are the same (but the third is different)

and all of the lunch boxes are different?

Same means indistinguishable and different means distinguishable.

Thanks!

 

Let candi 1 = sort a

Let candi 2 = sort a

Let candi 3 = sort b

 

\(\begin{array}{|c|c|c|c|c|} \hline & \text{Box $1$} & \text{Box $2$} & \text{Box $3$} & \text{Box $4$} \\ \hline 1& b & a & a & \\ 2& b & a & & a \\ 3& b & & a & a \\ 4& a & b & a & \\ 5& a & b & & a \\ 6& & b & a & a \\ 7& a & a & b & \\ 8& a & & b & a \\ 9& & a & b & a \\ 10& a & a & & b \\ 11& a & & a & b \\ 12& & a & a & b \\ \hline \end{array} \)

 

12 ways

 

laugh

heureka  Mar 6, 2018
 #2
avatar+995 
+4

Heureka, your answer is correct if a box is restricted to a maximum of one piece of candy.

nCr(4, 1) +  nCr(3, 2) = 12

Reasoning:

There are nCr(4, 1)  =  4 ways to place the unique piece of candy in one of the four (4) boxes.  Then the 2 indistinguishable candies can be placed in the other 3 boxes nCr(3, 2)  = 8 ways.  There are (4+8) = 12 ways to distribute the candies.

 

 

If there are no restrictions on the maximum then:

 

\(\hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\ \text {Reasoning:}\\ \text {All candy in 1 of 4 boxes (4 ways)}\\ \text {2 combinations (XX|Y & XY|X) distributed to boxes in (2)(4*3) = (2)(4!/2!) =(24) ways;}\\ \text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\ \text {Total 4+24+12 = 40 ways.} \)

 

(Adapted from Nauseated’s solution https://web2.0calc.com/questions/i-really-need-help-please-i-have-no-idea-how-to-do-it-sorry#r4)

 

 

GA

GingerAle  Mar 6, 2018
edited by GingerAle  Mar 6, 2018
 #3
avatar+19344 
+3

Thank You, GA.

 

Since no one answered my last question I'm going to post it again.
I have 3 pieces of candy to place in 4 lunch boxes.
In how many ways can I do this if exactly two of the candies are the same (but the third is different)
and all of the lunch boxes are different?
Same means indistinguishable and different means distinguishable.

 

Let candi 1 = sort a
Let candi 2 = sort a
Let candi 3 = sort b

 

\(\begin{array}{|c|c|c|c|c|} \hline & \text{Box $1$} & \text{Box $2$} & \text{Box $3$} & \text{Box $4$} \\ \hline 1& b & a & a & \\ 2& b & a & & a \\ 3& b & & a & a \\ \hline 4& a & b & a & \\ 5& a & b & & a \\ 6& & b & a & a \\ \hline 7& a & a & b & \\ 8& a & & b & a \\ 9& & a & b & a \\ \hline 10& a & a & & b \\ 11& a & & a & b \\ 12& & a & a & b \\ \hline \\ \hline 13& ba & a & & \\ 14& ba & & a & \\ 15& ba & & & a \\ \hline 16& a & ba & & \\ 17& & ba & a & \\ 18& & ba & & a \\ \hline 19& a & & ba & \\ 20& & a & ba & \\ 21& & & ba & a \\ \hline 22& a & & & ba \\ 23& & a & & ba \\ 24& & & a & ba \\ \hline \\ \hline 25& baa & & & \\ 26& & baa & & \\ 27& & & baa & \\ 28& & & & baa \\ \hline \\ \hline 29& aa & b & & \\ 30& aa & & b & \\ 31& aa & & & b \\ \hline 32& b & aa & & \\ 33& & aa & b & \\ 34& & aa & & b \\ \hline 35& b & & aa & \\ 36& & b & aa & \\ 37& & & aa & b \\ \hline 38& b & & & aa \\ 39& & b & & aa \\ 40& & & b & aa \\ \hline \end{array}\)

 

40 ways

 

laugh

heureka  Mar 6, 2018
edited by heureka  Mar 6, 2018

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