Before we move on to examples, let’s talk about all possible selections in the case when some or all of the objects are identical.
Let’s begin with the case when all of the objects are identical. Recall that the number of ways to select r objects out of n identical objects is 1.
Similar to what we did before, we have to select either no object, or select one object, or select two objects, and so on till selecting all objects. But in this case, selecting any number of objects can only be done in one way, instead of nCr ways.
That is, selecting no object can be done in 1 way.
Selecting one object can be done in 1 way.
Selecting two objects can be done in 1 way.
And, so on, selecting all the objects can be done in 1 way as well.
Using the addition principle again, the total number of ways to select any number of objects, or the total number of possible combinations will be 1 + 1 + 1 + … + 1 (n + 1 times) = n + 1
I think this was quite easy to understand. Now let’s move on to the case where only some of the objects are identical.
Suppose you have a lot of 10 objects out of which exactly 4 are identical and rest of them are different from each other (and also from the 4 identical ones). How many ways are there to select any number of objects from this lot?
We’ve got the tools for counting the case for both cases – when all of the objects are identical, and all of them are distinct. So all we have to do is separate the lot into two parts – one which has the 4 identical objects, and the other having the 6 different objects.
We’ll now divide our task of selection into two subtasks – make a selection from the identical lot, make a selection from the second lot, and multiply the number of ways. That’s it!
To select any number of objects from the first one (i.e. that containing the 4 identical objects) the number of ways will be 4 + 1 or 5.
To select any number of objects from the second lot (containing 6 different objects) the number of ways will be 26
Now, using the multiplication principle, the total number of ways is 5 x 26.
To convert this into formulas, the number of ways to select any number of objects from a lot of ‘p’ identical objects, and ‘q’ different objects will be (p + 1)2q
We can also extend the formula for the case where we have ‘p’ identical objects of one kind, ‘q’ identical objects of another kind, and ‘r’ different objects. Do you know the answer?
The required expression will be (p + 1)(q + 1)2r. I hope you got the idea.
Lesson Summary
- The number of ways to select any number of objects from a lot of n identical objects will be n + 1
- The number of ways to select any number of objects from a lot of ‘p’ identical objects of one kind, ‘q’ identical objects of another kind, and ‘r’ different objects will be (p + 1)(q + 1)2r
I think we have enough tools now to dive into problems. See you in the next lesson!