<s>[INST] You flip a coin $10$ times. How many ways can you get at least $7$ heads? 
You flip a coin $10$ times. How many ways can you get at least $7$ heads?

My answer.
$$\binom{10}{10}+ \binom{10}9\cdot\binom{10}1 + \binom{10}8\cdot\binom{10}2+\binom{10}7\cdot\binom{10}3$$
You have $10$ Heads and $0$ tails $+$ $9$ Heads $\cdot$ $1$ Tail $+$ $8$ Heads $\cdot$ $2$ tails $+$ $7$ Heads $\cdot$ $3$ tails.
The answer is $176$ though. 
 [/INST] the combinations notation ${10\choose 7} = \frac {10!}{7!3!}$ accounts for the fact that if you have 7 heads you also have 3 non-heads.  ${10\choose 7}{10\choose 3}$ is effectively squaring the value that you need for that term.  Similar for the other terms.
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