Up for the Challenge?

Find the number of different ways of distributing 30 identical objects into 3 identical boxes, so that each box contains at least one object.

NOTE- You cannot distinguish between the objects and the boxes. Also, (10,9,11) and (9,11,10) are one and the same . REMEMBER, the box and the objects are identical!


The answer is 75.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Andy Hayes
May 17, 2016

Relevant wiki: Identical Objects into Identical Bins

There are a number of identities on the identical objects into identical bins page that can be used for this.

p ( 30 , 3 ) = p ( 27 , 1 ) + p ( 27 , 2 ) + p ( 27 , 3 ) = 1 + 13 + p ( 27 , 3 ) = p ( 27 , 3 ) + 14 p(30,3)=p(27,1)+p(27,2)+p(27,3)=1+13+p(27,3)=p(27,3)+14

p ( 27 , 3 ) = p ( 24 , 1 ) + p ( 24 , 2 ) + p ( 24 , 3 ) = 1 + 12 + p ( 24 , 3 ) = p ( 24 , 3 ) + 13 p(27,3)=p(24,1)+p(24,2)+p(24,3)=1+12+p(24,3)=p(24,3)+13

p ( 24 , 3 ) = p ( 21 , 1 ) + p ( 21 , 2 ) + p ( 21 , 3 ) = 1 + 10 + p ( 21 , 3 ) = p ( 21 , 3 ) + 11 p(24,3)=p(21,1)+p(21,2)+p(21,3)=1+10+p(21,3)=p(21,3)+11

p ( 21 , 3 ) = p ( 18 , 1 ) + p ( 18 , 2 ) + p ( 18 , 3 ) = 1 + 9 + p ( 18 , 3 ) = p ( 18 , 3 ) + 10 p(21,3)=p(18,1)+p(18,2)+p(18,3)=1+9+p(18,3)=p(18,3)+10

p ( 18 , 3 ) = p ( 15 , 1 ) + p ( 15 , 2 ) + p ( 15 , 3 ) = 1 + 7 + p ( 15 , 3 ) = p ( 15 , 3 ) + 8 p(18,3)=p(15,1)+p(15,2)+p(15,3)=1+7+p(15,3)=p(15,3)+8

p ( 15 , 3 ) = p ( 12 , 1 ) + p ( 12 , 2 ) + p ( 12 , 3 ) = 1 + 6 + p ( 12 , 3 ) = p ( 12 , 3 ) + 7 p(15,3)=p(12,1)+p(12,2)+p(12,3)=1+6+p(12,3)=p(12,3)+7

p ( 12 , 3 ) = p ( 9 , 1 ) + p ( 9 , 2 ) + p ( 9 , 3 ) = 1 + 4 + p ( 9 , 3 ) = p ( 9 , 3 ) + 5 p(12,3)=p(9,1)+p(9,2)+p(9,3)=1+4+p(9,3)=p(9,3)+5

p ( 9 , 3 ) = p ( 6 , 1 ) + p ( 6 , 2 ) + p ( 6 , 3 ) = 1 + 3 + 3 = 7 p(9,3)=p(6,1)+p(6,2)+p(6,3)=1+3+3=7

p ( 12 , 3 ) = p ( 9 , 3 ) + 5 = 7 + 5 = 12 p(12,3)=p(9,3)+5=7+5=12

p ( 15 , 3 ) = p ( 12 , 3 ) + 7 = 12 + 7 = 19 p(15,3)=p(12,3)+7=12+7=19

p ( 18 , 3 ) = p ( 15 , 3 ) + 8 = 19 + 8 = 27 p(18,3)=p(15,3)+8=19+8=27

p ( 21 , 3 ) = p ( 18 , 3 ) + 10 = 27 + 10 = 37 p(21,3)=p(18,3)+10=27+10=37

p ( 24 , 3 ) = p ( 21 , 3 ) + 11 = 37 + 11 = 48 p(24,3)=p(21,3)+11=37+11=48

p ( 27 , 3 ) = p ( 24 , 3 ) + 13 = 48 + 13 = 61 p(27,3)=p(24,3)+13=48+13=61

p ( 30 , 3 ) = p ( 27 , 3 ) + 14 = 61 + 14 = 75 p(30,3)=p(27,3)+14=61+14=75

Thus, there are 75 \boxed{75} ways to distribute the 30 objects into the 3 boxes.

u can skip all of that and observe patterns

Leander Dsouza - 4 years ago

How much time (days) did it took you to write the latex code for it. Ha ha. But Andy sir if you have (you must have) written the wikis, for the various types of distribution, they are just absolutely BRILLIANT!

Prayas Rautray - 3 years, 9 months ago

The actual number of cases are 61 because no group should be left empty.

The Hindu - 1 year, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...