combinatorics – A coin is flipped 15 times. How many possible outcomes contain exactly four tails? contain at least three heads? – Mathematics Stack Exchange

$\begingroup$ Let ‘s try to come up with an theme for a ) :
We start with something easier : How many potential outcomes contain precisely one tails ? Well obviously $15$ since it can happen at either of these 15 throws .
Let ‘s update our interrogate to : How many potential outcomes contain precisely two tails ? We know the first tail has 15 possible throws where it can appear. The second tails has to happen at a different throw, so there are lone 14 throws left. But each of these throws would be approve. therefore what does this beggarly ? Well, for each of the 15 possibilities for the first tails there are 14 possibilities for the second tails left. We are tempted to say the answer is therefore $15\cdot 14 = 210$ but we would be ill-timed. There is a problem we need to fix first. We counted every possible combination doubly ! For exercise we could have placed the first tails at throw 1 and the second tails at throw 2 but we could besides have placed the first tails at throw 2 and the second at confuse 1. In the end this yields the same result, meaning beginning and irregular throw are tails and rest is heads. We can fix this by dividing by $2$. So we get the answer  \frac { 15\cdot 14 } { 2 } = 105\ ,. 

thus what happens if we ask : How many possible outcomes contain precisely three tails ? The theme is the same. We have 15 throws to place the first tails. We have 14 positions to place the second tails. immediately we are left with 13 throws to place the third base tails. Combining these we get $15\cdot 14\cdot 13$ possibilities which contain duplicates like ahead. We need to find out how many duplicates of each hypothesis we have. In our second example this was $2$. here it ‘s a bit more complicate. We can have first tails throw 1, moment tails throw 2 and third tails throw 3. nowadays we could besides have beginning tails throw 2, second tails throw 3 and third base tails throw 1. thus in perfume, for every possible way to arrange the three different tails at throw 1, 2 and 3, we have a twin. How many ways are there to arrange the first, second and third base tails ? That ‘s $3 \cdot 2 \cdot 1 =6$. It ‘s the same logic as ahead. The first buttocks has three positions to go to, the seoncd has two, the last has one .
This yields a recipe which we can now state of matter. It’s  \frac { normality ! } { thousand ! \cdot ( n-k ) ! } 
This rule calculates how many ways there are to take $k$ things from $n$ total things. In your case this would be how many ways there are to have $k$ many throws land on tails when performing a full of $newton$ throws.

A short admonisher :  normality ! = n \cdot ( n-1 ) \cdot ( n-2 ) \cdot \ldots \cdot 2 \cdot 1\ ,.  So for model  3 ! = 3 \cdot 2 \cdot 1 = 6\ ,. 
Let ‘s try to understand this convention. First look alone at  \frac { n ! } { ( n-k ) ! } \ ,.  Let ‘s put some numbers in. We chose $n=15$ and $k = 2$. We get  \frac { 15 ! } { ( 15-2 ) ! } = \frac { 15 ! } { 13 ! } = \frac { 15 \cdot 14 \cdot 13 \cdot 12 \cdot \ldots \cdot 2\cdot 1 } { 13 \cdot 12 \cdot \ldots\cdot 2 \cdot 1 } = 15 \cdot 14\, ,  because many things cancel out. comment that this is precisely the first thing we calculated in case 2. so this separate  \frac { normality ! } { ( n-k ) ! } = newton \cdot ( n-1 ) \cdot\ldots\cdot ( n- ( k-1 ) )  actually calculates this first measure in our examples. It tells us how many ways there are to take $k$ many ordered things from $normality$ full things.

We are not concern in orderings of $kilobyte$. So for us each manner in which we take the $kilobyte$ things from the same positions is the like, careless of ordering. We consequently have to calculate how many orderings of $potassium$ there are. And thats $kelvin !$, it ‘s precisely the argument we used in in exercise 3. Therefore we need to divide by $potassium !$. This yields the complete convention  \frac { nitrogen ! } { kelvin ! \cdot ( n-k ) ! } \ ,.