IIT JEE , AIEEE Forum - Mathematics , Algebra - Contest [swordfish #3]: Find ways to distribute identical balls in identical boxes

PORSHE

IMAGINE 30 DOTS &WE HAVE TO PUT 2BARS IN BETWEEN THEM

SO TOTAL NO OF WAYS 29C2(NO OF BALLS IN EACH BOX>0

Ronnieryan

ya porshe is correct
nice approach

btetarbe

Let us first fill all the 3 boxes with 3 identical balls which can be filled in 1 way.
Now if all the balls are necessary to be distributed then,
there is all in all ²⁷C₃ ways to distribute all the identical balls.

nitigya

we hv 30 identical balls and hv 3 identical markers(or boxes) as boXes should hv at least 1 ball ,the effective space for markers to be placed are 29

hence ²⁹C₃

celestine

IS THE ANS 3^29 /2 - 1

aryan.9001

arrange in a row 30 balls and calculate the no. of ways to put 2 partions in 29 places in between

Ans : ²⁹C₂

abhi3_raj

place one ball each in each of the boxes nd then distribute the other 27 randomly....the number of ways of doing this is 29C2 ways......but the boxes are identical so the final answer will become.......29C2/3

allamraju

First supply each box with one ball.The remaining 27 balls can be distributed by taking them along with 2 other identical things and arranging in a row in 29!/27!2!=29c₂ ways.Since the boxes are identical,The same partitions are considered as one.So,the no. of ways are 29c₂/3

ravi00

I think it is 30C3 and it will be uniform distribution and maximum no.of distribution.
plzzz..........corrct me if i m wrong........

ravi00

if we take equal distribution den d no.of arrangements will be 10x10x10=1000 ways

necrophilia666

56.

here's how.

a minimum of a ball per box. let box B and C have 1 ball each. then box A has 28. keep on decreasing a ball from box A and figure the combinations in boxes B and C. the total is 1+1+2+2+3+3+4+4+5+5+6+6+7+7.

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First supply each box with one ball.The remaining 27 balls can be distributed by taking them along with 2 other identical things and arranging in a row in 29!/27!2!=29c2 ways.Since the boxes are identical,The same partitions are considered as one.So,the no. of ways are 29c2/3

First supply each box with one ball.The remaining 27 balls can be distributed by taking them along with 2 other identical things and arranging in a row in 29!/27!2!=29c₂ ways.Since the boxes are identical,The same partitions are considered as one.So,the no. of ways are 29c₂/3