Contents
- 1 How many ways can you put balls in a box?
- 2 How many ways are there to put 4 balls into 3 boxes?
- 3 How many ways are there to distribute 5 distinguishable balls into 4 distinguishable boxes so that no box is empty?
- 4 What is the permutations of n balls?
- 5 How many ways can you select 4 pool balls from a tray of 16 balls?
- 6 How many ways are there to put $4$ balls into $3$ boxes given that the balls can all be distinguished but the boxes are not distinguished Thus for example putting all the balls in the first box is counted as the same outcome as putting all the balls in the second?
- 7 What’s the problem with putting balls into bins?
- 8 How to calculate the number of ways to eliminate a ball?
How many ways can you put balls in a box?
In addition, each box can take any number of ball. How many ways are there to place the balls into the boxes? There are only six ways.
How many ways are there to distribute k balls into n distinct boxes?
If we are supposed to distribute k distinct objects to n identical recipients so that each gets at most one, we cannot do so if k>n, so there are 0 ways to do so. On the other hand, if k≤n, then it doesn’t matter which recipient gets which object, so there is only one way to do so.
How many ways are there to put 4 balls into 3 boxes?
There are, again, three ways to arrange this one into the boxes. By the multiplication rule, we now have 3⋅6=18 ways to arrange the three balls into the boxes. We have two distinct balls, so we repeat the last step again, for 3⋅18=54 total ways to arrange these four balls.
How many ways can you put 5 balls in 3 boxes?
- 4 Answers. #1. +5. You can put the first ball in any of 3 boxes, the second ball in any of 3 boxes, etc. There are 3 choices for each ball, so there are 3x3x3x3x3 = 35 choices = 243 choices. geno3141 Mar 29, 2015.
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How many ways are there to distribute 5 distinguishable balls into 4 distinguishable boxes so that no box is empty?
Choose which two balls are to be paired, (52)=10, and then arrange the four objects, 4! =24, for a total of 240. Your first part is correct. The number of the total ways is 45.
How many ways are there of placing 4 distinct balls into 3 distinct baskets if each basket must have a ball?
19 ways… first select any 3 balls out of 4… no. of ways is 4 (4C3) …then the last ball can go into any of the three boxes…so no.
What is the permutations of n balls?
In permutation, objects are to be arranged in particular order. It is denoted by n P r or P(n, r)….Permutations & Combinations: Important Formulas.
| Number of permutation of these N balls = | N! |
|---|---|
| (B1)! × (B2)! × (B3)! × – – – – – (Br)! |
How many ways are there to distribute 6 distinguishable balls into 4 distinguishable boxes so that no box is empty?
If we let s be the number of ways to do this, then the number of ways to distribute 6 distinguishable objects into 4 distinguishable boxes so that no box is empty is given by 4! (s)=24s, since there are 4! ways to label the boxes.
How many ways can you select 4 pool balls from a tray of 16 balls?
= 4 × 3 × 2 × 1 = 24 different ways, try it for yourself!) So, our pool ball example (now without order) is: 16! 16!
How many ways are there to put 4 indistinguishable balls into 2 distinguishable boxes?
So, we have 1+4+6+4+1=16 ways to distribute the balls but since the boxes are indistinguishable we must divide by 2! (or just 2) to eliminate redundancy. So, 8 ways of arranging the balls in groups.
How many ways are there to put $4$ balls into $3$ boxes given that the balls can all be distinguished but the boxes are not distinguished Thus for example putting all the balls in the first box is counted as the same outcome as putting all the balls in the second?
There is no restriction of putting the balls in the boxes despite mentioning that there are 4 distinct balls and 3 distinct boxes. Now, the first box can get any of the 4 balls in 4 ways. Similarly, the second box can also get 4 balls in 4 ways and so is the third box. Hence, the total number of ways is 4*4*4 = 64.
How many are there to place 8 indistinguishable balls into four distinguishable bins?
How many ways are there to place 8 indistinguishable balls into 4 distinguishable bins? = 11!/(8!
What’s the problem with putting balls into bins?
The problem involves m balls and n boxes (or “bins”). Each time, a single ball is placed into one of the bins. After all balls are in the bins, we look at the number of balls in each bin; we call this number the load on the bin and ask: what is the maximum load on a single bin?
What do you call the number of balls in a bin?
Each time, a single ball is placed into one of the bins. After all balls are in the bins, we look at the number of balls in each bin; we call this number the load on the bin and ask: what is the maximum load on a single bin? Obviously, it is possible to make the load as small as m / n by putting each ball into the least loaded bin.
How to calculate the number of ways to eliminate a ball?
To ditribute the m k − n “elimination” into k bins, we have ( k m − n + k − 1 k − 1) different ways if m k − n ≤ m, i.e. the number of elimination in each bin will not exceed m. Thanks for contributing an answer to Mathematics Stack Exchange!
How to calculate the number of eliminations in each bin?
Suppose we start from the state that all bins are filled with m balls in each. Then the task for us is to eliminate m k − n balls from these k bins. To ditribute the m k − n “elimination” into k bins, we have ( k m − n + k − 1 k − 1) different ways if m k − n ≤ m, i.e. the number of elimination in each bin will not exceed m.