SOLUTION: 1. Five cards are chosen from a random a standard deck of playing cards. In how many ways can the cards be chosen under each of the following condition. a.) all are hearts. b.) Ex

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Question 1026751: 1. Five cards are chosen from a random a standard deck of playing cards. In how many ways can the cards be chosen under each of the following condition.
a.) all are hearts. b.) Exactly three kings
2. Use mathematical induction to prove 2+2+8+..+2^n= 2(2^n -1)
THANK YOUUUUU.
Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
1. Five cards are chosen from a random a standard deck of
playing cards. In how many ways can the cards be chosen
under each of the following condition.
a.) all are hearts. 
13 hearts, CHOOSE 5

Answer:  13C5 = %2813%2A12%2A11%2A10%2A9%29%2F%281%2A2%2A3%2A4%2A5%29 = 1287

[5 factors in the numerator, coming down from 15, 5 factors in the
denominator, coming up from 1.]

b.) Exactly three kings
Here we have two things to choose, 3 kings and 2 non-kings.

First we find the number of ways to choose 3 kings.

4 kings, choose 3.  4C3 = %284%2A3%2A2%29%2F%281%2A2%2A3%29 = 4 

Next we find the number of ways to choose 2 non-kings.

48 non-kings, choose 2 = 48C2 = %2848%2A47%29%2F%281%2A2%29 = 1128

Now for each of the 4 ways to choose the 3 kings, there
are 1128 ways to choose the 2 non-kings, so we multiply
those numbers:

4*1128 = 4512

---------------------------

Use mathematical induction to prove 2+2+8+..+2^n= 2(2^n -1) 
We are to prove P(n) which is

2%2B2%2B8%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2B2%5En=2%282%5En-1%29

First let's see what P(k+1) would be: [That's always the first 
thing to do.  Before you start an induction proof,  you should 
calculate P(k+1) to see where you're headed]: 

To do that, replace n by k+1 in 2%282%5En+-1%29 to see what P(k+1) 
is, for that is what we are going for, and if we have that 
beforehand, we'll know when we have arrived and the proof is 
finished. 

Substituting k+1 for n in 2%282%5En-1%29, we have 

2%282%5E%28k%2B1%29+-1%29 = 2%5E%28k%2B2%29-2

Now that we know what P(k+1) is, we know where we're going, and we'll 
know we have arrived if and when we get 2%5E%28k%2B2%29-2.  So now we 
can start the proof: 

P(1):  substitute n=1, 2%282%5E1-1%29 = 2, which is true, because the 
first term is 2. 

Assume P(k): 2%2B2%2B8%2B%22%22%2A%22%22%2A%22%22%2A%22%22%2B2%5Ek+=+2%282%5Ek-1%29  

We add the next term 2%5E%28k%2B1%29 to both sides:

%22%22=%22%22 
                       2%5E%28k%2B1%29-2%2B2%5E%28k%2B1%29%22%22=%22%22
                       2%2A2%5E%28k%2B1%29-2%22%22=%22%22 
                       2%5E%28k%2B2%29-2 


and now we see that we get the same P(k+1) as the one that we found 
in the beginning that we were going for.  So the proof is finished. 
So since P(1) is true, P(1) proves P(2), P(2) proves P(3), P(3)
proves P(4), etc., etc., ad infinitum. 

Edwin