<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 number of sequences of ten tosses that contain exactly seven heads and three tails is 
$$\binom{10}{7}\binom{3}{3} = \binom{10}{7}$$
since there are $\binom{10}{7}$ ways to select exactly seven of the ten positions for the heads and $\binom{3}{3}$ ways to select all three of the remaining three positions for the tails.  
By similar reasoning, the number of sequences of ten tosses that contain exactly $k$ heads and $10 - k$ tails is
$$\binom{10}{k}\binom{10 - k}{10 - k} = \binom{10}{k}$$
since there are $\binom{10}{k}$ ways to select exactly $k$ of the ten positions in the sequence for the heads and $\binom{10 - k}{10 - k}$ to select all $10 - k$ of the remaining $10 - k$ positions for the tails.
Hence, the number of sequences of ten coin tosses in which at least seven heads occur is 
\begin{align*}
\sum_{k = 7}^{10} \binom{10}{k}\binom{10 - k}{10 - k} & = \sum_{k = 7}^{10} \binom{10}{k}\\
& = \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10}\\
& = 120 + 45 + 10 + 1\\
& = 176
\end{align*}
In your attempt, when you calculated the number of ways of selecting seven heads and three tails, you first selected seven of the ten positions for the heads and then selected three of the ten positions for the tails without taking into account that the three tails can only occupy the $10 - 7 = 3$ positions not already occupied by heads.  That means you allowed heads and tails to occupy the same positions in the sequence, a physical impossibility.
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