In this part, I will be covering the remaining three types of questions from the topic of distribution of alike objects.
In the earlier part, we saw how to solve questions involving distribution of alike objects when the number of positive, or non-negative integral solutions are to be found. We also saw the number of ways of distribution when one or more of the coefficients of the variables is not unit in nature. Let’s look at a few, trickier fresh cases.
Less than n
It essentially means that the expression will be less than a particular number. For example: There are less than 10 Buzz Lightyear action figures to be distributed among three children. In how many ways can they be distributed?
It translates to:
a + b + c < 10
where a, b and c are non-negative integers. This means that the number of action figures is unknown but it is a non-negative integer less than 10. It involves all the cases starting from distribution of 0 action figures to 9 action figures. While it would give you the same answer if you calculate for individual cases and add them up, we can use a shorter method to save us some precious time.
To solve these type of questions, you have to simply put in another variable and turn the inequality into an equation.
a + b + c + d = 10
Now, d has to be greater than or equal to 1 for it to hold true. Remember that we cannot have 10 toys and so, cases involving d = 0 are moot. So, d has to be a positive integer. In other words, we are simply saying that there are 4 kids and we have 10 action figures to be distributed such that d gets at least 1 action figure.
To account for this nature of d, we will have to remove 1 from the RHS as we saw in the previous article. The new expression becomes
a + b + c + d = 9 where a, b, c, d > 0
The number of solutions to this is 12C3 = 220 ways.
Alternatively, you could have simply written the original inequality as
a + b + c + d = 9 where d > 0
Please note: If you had proceeded using individual cases, you would have got
a + b + c = 0, 2C2 = 1 solutions
a + b + c = 1, 3C2 = 3 solutions
a + b + c = 2, 4C2 = 6 solutions
a + b + c = 3, 5C2 = 10 solutions
a + b + c = 4, 6C2 = 15 solutions
a + b + c = 5, 7C2 = 21 solutions
a + b + c = 6, 8C2 = 28 solutions
a + b + c = 7, 9C2 = 36 solutions
a + b + c = 8, 10C2 = 45 solutions
a + b + c = 9, 11C2 = 55 solutions
Total of 220 solutions.
Less than or equal to n
This is extremely similar to the above case. Let’s see a question. There are at most 10 Buzz Lightyear action figures to be distributed among three children. In how many ways can they be distributed?
In this case, you have to figure out the number of ways in which to distribute 0 action figures, 1 action figure, 2 action figures and so on till 10 action figures. Mathematically, you have to find the number of non-negative integral solutions to
a + b + c < 10
Introducing another variable, we get
a + b + c + d = 10 where, d > 0
Total number of solutions is 13C3 = 286 ways.
Special variables
We saw that there could be variations when it comes to coefficients. But, in some cases, the variable might have some special conditions that need to be fulfilled.
For example: There are 45 Buzz Lightyear action figures to be distributed among five children. In how many ways can they be distributed such that each kid gets an odd number of action figures?
For those of you who are not initiated, it would be extremely difficult to solve this question. You have to take variables such that your solution set will have odd numbers only.
A positive odd number can be represented in terms of 2x + 1 where x belongs to the set of whole numbers. So, we will take the variables as
2a + 1, 2b + 1, 2c + 1, 2d + 1, 2e + 1
So, the equation becomes
2a + 1 + 2b + 1 + 2c + 1 + 2d + 1 + 2e + 1 = 45
2a + 2b + 2c + 2d + 2e = 40
a + b + c + d + e = 20
Number of ways of distribution is 24C4.
You can try out another question similar to the one we just solved. There are 50 Buzz Lightyear action figures to be distributed among five children. In how many ways can they be distributed such that each child gets an even number of toys?
Positive, unequal integral solutions
This is probably the toughest question that can pop up from this topic. The very mention of the word ‘unequal’ should be enough to let go of this question type in the first go. Let’s see an example. There are 10 Buzz Lightyear action figures to be distributed among three children. In how many ways can they be distributed such that each kid gets at least one action figure and no two kids get an equal number of action figures?
The easiest way to solve this is to give 1 action figure to a, 2 action figures to b and 3 action figures to c. This is because we are assuming that a gets lesser action figures compared to b who in turn gets lesser action figures compared to c. So, we are left with 4 action figures to be distributed among 3 kids.
a + b + c = 4
The important step here is to understand that the values of a, b and c have to be such that a < b < c. This is because, we do not want to offset the assumption that we made when we gave 1 to a, 2 to b and 3 to c. If this is not followed, we could end up with a case where a gets more action figures than b which leads to a violation of the assumption.
The solution sets to a + b + c = 4 such that c > b > a will be: (0, 0, 4) (0, 1, 3) (0, 2, 2) (1, 1, 2) so, 4 solutions.
Similarly, we will have 4 solutions for each of a>c>b, b>a>c, b>c>a, c>a>b and c>b>a. So, total of 4*6=24 ways of distribution.
For the technically inclined, here are the solution sets:
| 1,2,7 | 7,1,2 | 3,1,6 | 1,4,5 | 5,1,4 | 3,2,5 |
| 1,7,2 | 7,2,1 | 3,6,1 | 1,5,4 | 5,4,1 | 3,5,2 |
| 2,1,7 | 1,3,6 | 6,1,3 | 4,1,5 | 2,3,5 | 5,2,3 |
| 2,7,1 | 1,6,3 | 6,3,1 | 4,5,1 | 2,5,3 | 5,3,2 |
Mixed cases
There could be questions which would have a combination of the concepts explained above. These have to be dealt with similarly to how the parent cases were solved. You can try out a few questions to get more comfortable with the logic behind such questions.
There are less than 10 Buzz Lightyear action figures to be distributed among three children. In how many ways can they be distributed such that the second child gets an even number of action figures?
There are at most 10 Buzz Lightyear action figures to be distributed among three children. In how many ways can they be distributed such that the first child gets a multiple of 3 action figures?
These are the only question types that can appear from this area. I hope that at least a few of the doubts that you had with regard to distribution of alike objects have been addressed. Let us know in case you want an in-depth analysis of a topic and we will try to cover it in the coming articles.
All the best!
