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.
On
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.
On
Let's look at $\binom{10}{7} \cdot \binom{10}{3}$. This counts the number of ways you can
This includes, for example, choosing the label the first seven coins as "heads" and the first three coins as "tails".
This choice of labels cannot possibly describe a result of flipping a coin ten times, because three coins have both labels and three coins have neither label.
The simple fix is to recognize that all you need to do is to count the number of ways you can
Once you've done this, you can then proceed to label the three remaining coins as "tails".
In this system, the possible choices for labels are in bijective correspondence with the possible ways to get seven heads in ten coin flips.
On
The problem is that once you have chosen which $9 $ of the ten flips are going to be heads, there's is no choice remaining as to which one is a tail. It has to be the last one not selected.
Hence the formula starts with $\ldots +{10\choose 9 }$, times nothing (or times $1\choose 1 $ if you will).
On
Consider thinking 10 flipping of a coin and getting n<10 heads as putting n black balls(assume tails are white balls) in 10 boxes... Now, the question transforms to how many ways can you put at least 7 black balls in 10 boxes?
The answer is 10c10+10c9+10c8+10c7=176 . The permutation of white balls doesn't matter as interchanging the box no of white balls counts as a same result.....
On
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.
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 $=\dbinom{10}{7}+\dbinom{10}{8}+\dbinom{10}{9}+\dbinom{10}{10}=176$