SOLUTION: Compute each of the following. Look for simplifications first.
a. 20P15 (the 20 and the 15 are small)It's looking for the permutation??
b. (n+1)!
------
(n-1)!
T
Algebra.Com
Question 246255: Compute each of the following. Look for simplifications first.
a. 20P15 (the 20 and the 15 are small)It's looking for the permutation??
b. (n+1)!
------
(n-1)!
Thank you so much in advance.
This is so difficult.
~Marney
Found 2 solutions by jim_thompson5910, TheProdicalSon:
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
a)
Start with the given formula
Plug in and
Subtract to get 5
Expand 20!
Expand 5!
Cancel
Simplify
Now multiply 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6 to get 20,274,183,401,472,000
================================================================
b)
... Start with the given expression.
... Expand the numerator. Remember that n! = n(n-1)(n-2)(n-3)...(3)(2)(1). So (n+1)! = (n+1)(n+1-1)(n+1-2)(n+1-3)(n+1-4)...(3)(2)(1)
... Combine like terms.
... Take note that (n-1)! = (n-1)(n-2)(n-3)...(3)(2)(1) which is what the numerator (minus the first two terms) looks like. So rewrite (n-1)(n-2)(n-3)...(3)(2)(1) as (n-1)!
Note: if you're asking 'Why did we just do that?' The goal is to cancel out the factorials. Since the denominator has a (n-1)! term, we just need that term in the numerator for it to cancel out.
... Cancel out the common terms.
... Rearrange the terms.
... Distribute
So
where 'n' is an integer and
Answer by TheProdicalSon(34) (Show Source): You can put this solution on YOUR website!
The first one is pretty simple.
20P15 = 20(amount to choose from)permutation 15(amount needed)
this is written as which translates into (after simplification)
20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * 12 * 11 * 10 * 9 * 8 * 7 * 6
= 20,274,183,401,472,000
The second one.
I have no idea, I believe the answer is already as simplified as it can get
but I really don't know, sorry.
Comment on my page if this was helpful.
Sincerely,
TheProdicalSon
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