IIT JEE , AIEEE Forum - Mathematics , Algebra

just_4u

the number of ways in which 9 identical balls can be placed in 3 identical boxes ?

rajsh

the answer is 3 to the power 9
and i will give you a short trick to solve such type of problems
see
take the example of your question the ball is in the box means the ball is above and the box is under the ball
so the box taken as base and the ball is taken as exponent

i am giving you another example for a ring and a finger
in how many ways 3 rings can be weared in 5 fingers
so the answer is
5 to the power 3
bcz the the ring is above and the finger is under it
so finger is taken as base and the ring is taken as its power

if you like the answer plz plz rate me according to your thinking about the answer

lazycol

i dont think dat 3⁹ is ans

becoz balls & boxes r identical here,dis works only with distinct balls & boxes

possibilities r

9 0 0 ; 8 1 0; 7 1 1; 6 2 1; 5 3 1; 5 2 2; 4 4 1; 4 3 2; 3 3 3;

my ans = 9

asmit

yeah lazycol is correct since no restriction has been imposed on the number of balls in one box so logically 1st ball can be kept in the 3 boxes in 3 ways,same with 2nd,3rd ....so total ways is 3x3x3x3x....=3^9 ways

Oldisgold

A. Let the no of Balls be 'n' and no of boxes be 'x'.
Then the ans is (n+x-1) c n
ie in this case, (9+3-1) c 9 = 11c9 = 55

B. The logic by construct is as below.
1. for eg take 3 identical balls over 2 boxes. the no of combinations is : (3,0), (2,1), (1,2), (0,3). [ (3+2-1) c 3 ] = 4c3 = 4
2. Similarly take 4 over 3 boxes. the no of combo is : (4,0,0),(0,4,0),(0,0,4), (3,1,0),(3,0,1),(0,1,3),(0,3,1),(1,3,0),(1,0,3),(2,2,0),(2,0,2),(0,2,2),(2,1,1),(1,2,1),(1,1,2) = 15 [ (4+3-1) c 4] = 6c4 = 15.

C. The conceptual logic is like placing (3-1 = 2) sticks to seperate ' n ' balls. Since balls are identical, the only distinct combinations can be in terms of the distinct groupings of these balls in various boxes.. Rgds. Balaji TV

priyesh

see

since the balls & boxes are identical so we are not bothered about which ball goes into which box.We are only going to consider the no. of balls going into each box. & since boxes are identical hence arrangements such as 1,2,5 & 5,2,1 will be same

now suppose no. of balls going into first,second & third boxes are x₁, x₂, x₃ respectively

then x₁ + x₂ +x₃ = 9

& furthermore (x1,x2,x3) this pair of three should not rearrange

count the no. of solns. using this two conditions to get the ans.

avinash.sharma

i agree with Lazycool

joyfrancis

Take an example as 2,3,4 {9 balls in 3 boxes 2 in 1st, 3 in 2nd , 4 in third}, now imagine the boxes are kept in triangular format {just to make things more clearer}, .: the arrangements such as 4,3,2 and 3,2,4 etc would look the same because of the identical boxes.

.: We will count 2, 3 , 4 as only 1 possible arrangement and not 3!.

Now let us see how many possibilities are there

First case: there is 0 balls in one or more boxes.

Total cases are 0-0-9 , 0-1-8- , 0-2-7, 0-3-6 , 0-4-5

Second case: There is 1 ball in any 1 or more boxes.

Here we will exclude those cases in which we have 1&0 in two boxes since that is already counted above as 0-1-8

.:Possibilities are 1-1-7 , 1-2-6 , 1-3-5 , 1-4-4

Third case : There are 2 balls in 1 or more boxes.

Possibilities are : 2-2-5 , 2-3-4 (excluding 1-2-6 and 0,2,6 since they have already been counted)

Fourth case : There are 3 balls in 1 or more boxes

Possibilities are : 3-3-3

Only these many possibilies are there if we go futhur into 4, 5.. balls we will get arrangements which are already counted.

So the answer is 12 ways

Oldisgold

Sorry to be a bit of a wet blanket. But lazycol's solutions can still be expanded upon.

for eg. along with (7,1,1) , (7,2,0) can also be a combo. similiary, (6,2,1) can co exist with another combo (6,3,0).

Besides, the boxes may be identical. but there is a sequence to the boxes. hence, (9,0,0) is not the same as (0,9,0) or (0,0,9).

Hence, sorry to be a bit of a bother. but stick to my number 55 as the answer. rgds.

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