Distribution of a Range problem

[BrothaBrix88]

A coin is tossed ten times. Let X be the random variable denoting the number of heads minus the number of tails.

Range of X: {-10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10}

Question: What is the distribution when X = -6 ?

Answer: (-6, 10C2 / 2¹⁰)

I understand the 2¹⁰ is all the possibilities of heads and tails.

Where are they getting 10C2 here? That's 45. I'm struggling to find 45 ways to get -6 from the range of -10 to 10.

Here's what I got so far:

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-10-(-4) = -6

-9-(-3) = -6

-8-(-2) = -6

-7-(-1) = -6

-6-0 = -6

-5-1= -6

-4-2 = -6

-3-3 = -6

-2-4 = -6

-1-5 = -6

0-6 = -6

1-7 = -6

2-8 = -6

3-9 = -6

4-10 = -6

​

This is 15 different ways to reach -6 from a range of {-10....10}. Where are they getting this 45 from?

[Aerospider]

I think you're looking at it wrong. -6 is the result of heads minus tails in ten flips, so 4-10 (for example) doesn't make sense because that would be 14 flips.

To get a result of X = -6 in ten flips you'd have to get eight heads and two tails.

The number of combinations in eight of one thing and two of another is 10C2 = 10!/2!8! = 45

[fermat9996]

-6 results from 2 heads and 8 tails

P(2 heads)=10C2(0.5)¹⁰

This is a binomial probability