This lesson will cover briefly a few simple cases involving selections (or combinations) involving identical objects.
Suppose you have five identical red balls, of which you’ve to choose any two. How many different combinations are possible?
It doesn’t matter which two balls you select – whether you select the first two, or the last two, or the third and the fourth – you’ll end up with the same selection each time. Why? Because the balls are identical.
Therefore the number of ways to select two balls will be one.
And if you had to select three balls, how many choices would you have? The answer remains the same – one.
And four balls? Again, only one possible selection.
Okay, that was simple to understand. Let’s complicate things a bit now.
Suppose you now have the following lot – three red balls and three balls of different colors
If you’re to select three out of these, how many different choices do you have?
The answer again will not be 6C3 (the case where all of the balls are of different color), because identical balls will reduce the number of choices, as we saw above.
So how can we count the possible combinations in this case?
We first separate the balls into two lots – the identical balls (say, lot 1) and the distinct balls (lot 2).
Next, we divide our selection into two sub-tasks – select from lot 1 and select from lot 2.
Finally, we make cases..
I. All the three balls from lot 1: 1 way
II. Two from lot 1 and one from lot 2: 1 x 3C1 ways
III. One from lot 1 and two from lot 2: 1 x 3C2 ways
IV. All the balls from lot 2 – 3C3 ways
.. and we add all the cases: 1 + 3 + 3 + 1 = 8 ways
Hmm. It’s getting complicated already.
And you must be thinking, “Wait.. what about the case where we’ve to select 3 balls from a lot of 4 red, 5 green and 6 different colored balls?”
I’ll cover these and many other cases much later, after we’ve gathered all the tools.
- The number of ways to select r objects out of n identical objects equals 1 (where 0 ≤ r ≤ n)
See you in the next lesson, where I cover all possible combinations.