# Binary String's set of elements

Example: How many 7-digit binary strings have three 1's? Answer: $${ 7 \choose 3} = 35$$

definition of $${ n \choose k}$$ : If n and k are integers, then $${ n \choose k}$$ denotes the number of subsets that can be made by choosing k elements from an n-element set.

According to the above definition, $${ 7 \choose 3} = \frac{7!}{3!4!} = 35$$

My difficulty: in the example we're choosing 3 elements from from a 7-element set, however what is this 7-element set?

I thought about this because when we calculate 7! , it means we have some set of 7 elements, and in this example's case we have a binary string, which is composed of the set { 0 , 1 } , i.e. a 2-element set and not a 7-element set.

• Isn't $\frac{7!}{3!4!} = 35$? – Ak. Jan 25 at 10:45
• Yes, I edited calculation error.. – hazelnut_116 Jan 25 at 10:56

You can think of it this way: Your seven-element set is $$\{1, 2, 3, 4, 5, 6, 7\}$$, corresponding to the indices of the digits which form your string (i.e., $$3$$ corresponds to the third digit).

Now you choose three elements out of this set, corresponding to the spots of your string where a "one digit" should appear. (This is a binary decision, meaning that the other elements are "zero digits".)

The $$7$$ elements are the terms in the $$7$$-tuple that form the $$7$$-digit string.

Each of these elements is distinct.

The "choice" part of the exercise is: "Is this element a $$1$$ or a $$0$$?"

Hence this exercise is equivalent to the exercise of choosing $$3$$ elements from a $$7$$-element set.

It is "how many ways can I choose $$3$$ elements of this string of $$7$$ digits so that these $$3$$ elements are $$1$$ and all the others are $$0$$?"

And of course the answer is $$\dbinom 7 3 = \dfrac {7!} {3!4!} = \dfrac {5040}{6 \times 24}$$ which works out to be $$35$$.

• But why the elements of the 7 tuple are distinct? each of them can have only 2 possible values ( 0 and 1 ) – hazelnut_116 Jan 25 at 11:02
• @hazelnut_116 Because there are 7 of them. The "distinctness" is in their position. The first one, the second one, the third one, ..., the seventh one. – Prime Mover Jan 25 at 11:18

No.

$${ 7 \choose 3} = \frac{7!}{3!4!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}{(3 \cdot 2)(4 \cdot 3 \cdot 2)} = 7 \cdot 5 = 35.$$

• I miscalculated, I edited the question. – hazelnut_116 Jan 25 at 10:57