How many ways can select 3 letters from 6 letters (6 letter each letter is duplicate)

combinationscombinatoricspermutations

[A,A,B,B,C,C]

So we have the above 6 letters and we need to create a combination of 3, such that each letter will appear once in the result, just for the sake of understanding i name first A as A1, second A as A2 similarly for B and C.

A1 B1 C1
A1 B1 C2 
A1 B2 C1
A1 B2 C2 

A2 B1 C1
A2 B1 C2 
A2 B2 C1
A2 B2 C2

So in total there are 8 combinations,

I have two approaches to solve this,

Approach 1 :

We have three places => _ _ _

At First place out of 2'A we have a choice of selecting 1, So 2C1

and

At Second place out of 2'B we have a choice of selecting 1, So 2C1

and

At Third place out of 2'C we have a choice of selecting 1, So 2C1

So in total 2 * 2 * 2 = 8

Approach 2 :

Since we know the combination formula which is nCr = n!/r!(n - r)!

If we apply that 6!/3!3!

And since we know A,B and C are repeated twice we do like this : 6!/3!3!2!2!2!

My Question is why approach 2 is wrong ?

Best Answer

We wish to find how many different sets of the form $\{A, B, C\}$ we can form given two distinct $A$s, two distinct $B$s, and two distinct $C$s.

Since we have two choices for each of the three letters, there are indeed $2 \cdot 2 \cdot 2 = 8$ ways to form the set.

In a set, the order does not matter. For instance, $\{A, B, C\} = \{B, A, C\}$. In your second approach, you considered positions, so you were counting sequences rather than sets.

The expression $$\binom{6}{3} = \frac{6!}{3!3!}$$ is the number of ways we could obtain a sequence with three heads and three tails since choosing three of the six positions in the sequence for the heads completely determines the sequence since choosing positions of the three heads means the remaining positions must be filled with tails.

Similarly, the expression $$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{2!4!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{6!}{2!2!2!0!} = \frac{6!}{2!2!2!}$$ counts the number of sequences of length six with two indistinguishable $A$s, two indistinguishable $B$s, and two indistinguishable $C$s since we must select two of the six positions for the $A$s, two of the remaining four positions for the $B$s, and then must fill the remaining two positions with the $C$s.

It is not clear why you were using the combinations formula. The expression $$\binom{n}{k}$$ counts the number of subsets of $k$ elements which can be selected from a set of $n$ objects. In the expressions above, I chose subsets of positions in a sequence in which to place indistinguishable objects. However, your six objects are distinct.

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