63 views
How is the problem..

Distribute 5 toys such that each of 3 child get atleast 1

Different from sum of 3 no. X+y+z=5 such that each digit >= 1.

Plz explain ?
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+1
150 for first and 6 for the second?
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How do we need to approach both problems differently plz give some insight..

Do we need to take candies to be distinct or how do we approach?
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1. For the first one, I have considered toys to be different. When objects to be distributed are greater than the available persons and each should get at least one, we cannot do give one to each and then permute remaining anyhow. Here comes the concept of grouping and distribution.

2. The second one is the permutation of star and bars with extra criteria of having at least one from each. Take one from each which makes the equation as X' + Y' + Z' = 2 (Two stars and two bars)
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actually, it is questioner responsibility to give toys are identical or not ?
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Question will give the hint for the same
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unfortunately, i didn't get it from the question
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This way for the first case we can manually find the cases but how do we do if the no. To be distributed is large say 12
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shaik bro I meant the questions in the exam will give the hint:). This one is ambiguous.
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@Manoj Kumar Pandey refer Rosen or do combinatorics from here.

don't over think

this can be done  manually

give all student one toy first

x= 1

y=1

z=1

now you are left with 2 more toys i think you can distribute 2 among three easily

x               y              z

2              0              0

0              2              0

0              0              2

1              1              0

1              0              1

0              1              1

ANS=6

correct me if I am wrong.....
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