In how many ways 6 distinct balls can be distributed amoung 3 identicall urns ?please explain the answer ?

Answer 1

2^6 ways

Answer 2

This question is one pertaining to combination. If you call the 6 balls respectively a, b, c, d, e and f, the question is what is the arrangement problem. This question is more tough then the usual combination problems: all the balls can be in 1 urn while the others are empty. But according to your question, the balls must be all in some urn in all distributions considered.
For the 1st urn, the fact is simple: there exist 7 possible states:the urn can be empty, have 1 ball, 2...6. But for that facts sake, the other urns must be filled in function of the first. Ex: if the 1st urn has 1 ball in it, the second may have 2 (up to 5) and the 3rd one MUST have 3.

Since each of the choices of the numbers of the balls in the 1st urn changes the number of possible distributions of the balls in the other urns, this may need a little complicated operations.
For the 1st urn there are 7 choices, for the 2nd there are(after deciding for the first) a variable number of choices. Ex: if the 1st choice is that the there are no balls, in the second one there will be 7 choices. On the contrary, if the choice for the 1st urn is that it will have a set of 1 ball, the choice for the 2nd urn is reduced to 6 possible decisions. And, worse, if the choice for the first urn is that it will have all 6 balls, then the possibles choices for the 2nd and 3rd will be one choice: that of having 0 balls.
If you waste your time noting all the possible combinations on a sheet of paper you will never finish. You better use formulas.

I haven't read this formula anywhere. You have caused me to write it by asking that question:

There are a total of 7 possible choices for the first urn.

Choosing 6 gives 1 possible choice for the other positions.
Choosing 0 gives 7 possible choices for the other positions.
Choosing 4 gives 3 possible choices for the other positions.

When you look at these results correctly, you can see that the number of possible choices for each decision for the 1st urn is equal to 7 minus the chosen number for the 1st urn. Ex: The choice of 5 gives 2. You can write this formula as our old buddies the linear equations of 2 unknown numbers:
y = -x +7
But that doesn't help you much would you say. You'll change your mind when you realise that to solve the problem you are proposing, you need to follow a pattern. This could have been written in sigma notation, but I will simply say what you have to do:
You add the results given by all the numbers from 1 to 6 ( or 7, you'll get the same results), I mean after obtaining the results of the numbers when they replace x.
You will obtain (7-0)+(7-1) + (7-2) + (7-3) + ( 7-4) + (7-5) + (7-6) +(7-7)=7+6+5+4+3+2+1+0=28
So they are 28 possible ways to distribute 6 distinct balls amoung 3 identical urns.

Answer 3

it is very easy!
6 distinct balls
first of all can be distributed in 6 ways.
in the second turn u can distribute in 5 ways becoz remeber the ways hav to be distinct and u can't repeat the type of distribution u did in the first turn becoz the ways hav to be distinct and since u are distributing the balls among 3 urns so the no. of turns are 3.
then in the 3rd turn u can distribute by 4 ways since u hav already used the other 2 ways in the 1st and 2nd distribution.

thus total no. of ways= 6 x 5 x 4 =120 ways
u will know more about such probs when u consult ur maths teacher and ask him about permutationsand combinations

Answer 4

this is equal to no ways of 3^6into three factors
3^6=3^a*3^b*3^c
a+b+c=6
when a=0 b+c=6
and we get 06,15,24,33.we neglect others because remaining are identical
a=1 b+c=5 for this 05,14,23.
a=2 b+c=4for this 04,22,13.
a=3 b+c=3for this 03,21.
a=4 b+c=2 for this 02,11.
a=5 b+c=1for this 01.
a=6 b+c=0 for that 00.////total no of ways =4+3+3+2+2+1+1=16

Answer 5

the answer is
( 6+3-1)x(6+3-2)/2 =28

Answer 6

In 18 ways. Because the six balls are distinct each of them could get any of the balls six different times

Answer 7

6-0-0, 3
5-1-0, 6*6C5
4-2-0, 6*6C4
4-1-1, 3*6C4
3-3-0, 3*6C3
3-2-1, 6*6C3*3C2
2-2-2, 6C2*4C2
n = 3(1+2*6+3*15+20+2*20*3+6)
N = 3(1+12+45+20+40+120+6)
n=732

Answer 8

it will be solved by combination method
use formula n!/{r!*(n-r)!}
here n is 6
and r is 3
so , 6!/{3!*3!}
= 6*5*4/3!
=20