<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] Since we need at least $7$ heads from $10$ trails
First we get $7$ heads and $3$ tails in $\dbinom{10}{7}$
Second we get $8$ heads and $2$ tails in $\dbinom{10}{8}$
Third we get $9$ heads and $1$ tail in $\dbinom{10}{9}$
Fourth we $10$ heads and $0$ tails in $\dbinom{10}{10}$
Now total number of permutations $=\dbinom{10}{7}+\dbinom{10}{8}+\dbinom{10}{9}+\dbinom{10}{10}=176$
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