Question 1182155
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Part (a)


Let's label the five slots as A,B,C,D,E.
Each slot has the option of heads (H) or tails (T)
One such configuration could be HHHTT.
We could also have HHTHT
Each case involves three H's and two T's.


The question is: how many ways can we rearrange those sequences of H's and T's?


Let's rephrase the problem a slightly different way. Let's say we have 5 slips of paper, and each paper has a single letter A through E on it. 


If we randomly picked out slips of paper, then we might get this sequence:
AEB,CD
The first three letters would tell us which slots get H and the remaining two letters get T
So that sequence would then tell us we got HHTTH


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So we need to find the number of ways to select three letters from A,B,C,D,E. The order doesn't matter.


That's a combination problem involving the nCr formula. We plug in n = 5 and r = 3. You should get 10 different ways.


Another approach you could take is to realize that there are 5*4*3 = 60 different permutations possible and 60/(3!) = 60/(3*2*1) = 60/6 = 10 combinations. 


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We found there are 10 ways to get exactly three heads where the order doesn't matter. 


This is out of 2^5 = 32 ways to flip five coins. 



Therefore,
10/32 = (2*5)/(2*16) = <font color=red>5/16</font> is the final answer for part (a).


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Part (b)


This part is basically the same as part (a), but the numbers are different of course.


We have n = 8 coins and we want r = 2 heads. I'll use the nCr combination formula to show a different approach.


nCr = (n!)/(r!*(n-r)!)
8C2 = (8!)/(2!*(8-2)!)
8C2 = (8*7*6!)/(2!*6!)
8C2 = (8*7)/(2!) ...... a pair of 6! terms cancel
8C2 = (8*7)/(2*1)
8C2 = 56/2
8C2 = 28


We have 28 ways to get exactly 2 heads. The number 28 is found in Pascal's triangle in the row that reads 1,8,28,...


There are 2^n = 2^8 = 256 ways to flip 8 coins


The probability we want is 28/256 = (4*7)/(4*64) = <font color=red>7/64</font> which is the final answer.


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Part (c)


I'll use yet another approach. I mentioned Pascal's triangle earlier so why not go over an example of it.


Look at the row that has 1,8,28... in it. At the very end of the row, you should notice the same exact pattern except that it's backwards. 


Start at the very far right end (1). That corresponds to slot 8. Move one spot to the left and you'll get to 8, which is for slot 7. Slot 6 will get us to 28. 


This means that nCr = 8C6 = 28. The triangle has nice symmetry we can quickly determine combination values from.


Why does this work? Well the idea of getting 6 heads is basically the same as getting 2 tails. If we want to know how many exact ways to getting 2 tails, then we just follow the steps done in part (b) earlier and we should get 28 ways. The labels "heads" and "tails" don't really matter since either side is equally likely. We can easily flip the labels and get the same thing.


So that's why <font color=red>7/64</font> is the answer here, and it's not a coincidence that parts (b) and (c) have the exact same answer. 


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Part (d)


You should find that there are...
nCr = 4C2 = 6 ways to get exactly two heads
2^n = 2^4 = 16 ways to flip four coins.


Therefore, 6/16 = (2*3)/(2*8) = <font color=red>3/8</font> is the probability we want.


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Part (e)


Same story as before, but different numbers
nCr = 20C10 = 184,756 ways to get ten heads
2^n = 2^20 = 1,048,576 ways to flip twenty coins


We end up with the fraction (184,756)/(1,048,576)
aka 184756/1048576

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