Question 119877This question is from textbook Glencoe Algebra 2
: Solve for n.
35. P(n,4) = 40[P(n-1,2)]
36. P(n,4) = 3[P(n,3)]
37. n[P(5,3)] = P(7,5)]
38. 208P(n,2) = P(16,4)
how exactly do you solve for n? please and thank you.
This question is from textbook Glencoe Algebra 2
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Solve for n.
35. P(n,4) = 40[P(n-1,2)]
Using definition of P(n,r)= n!/(n-r)! you get:
n!/(n-4)! = 40[(n-1)!/(n-3)!]
Keep in mind n! = n(n-1)! and (n-3)!=(n-3)(n-4)!
Cancel where you can to get:
And you get:
n = 40[1/(n-3)]
Cross-multiply to get:
n^2-3n-40=0
Factor to get:
(n-8)(n+5)=0
Solve for n (Keep in mind that n must be positive).
n = 8
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36. P(n,4) = 3[P(n,3)]
Use the definition of P(n,r)
n!/(n-4)! = 3[n!/(n-3)!]
Cancel n! of both sides.
Remember (n-3)!= (n-3)(n-4)!, so cancel (n-4)! of both sides to get:
1/1 = 3/(n-3)
Cross-multiply to get:
3 = n-3
Solve for n.
n = 6
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37. n[P(5,3)] = P(7,5)]
Apply the definition of P(n,r) to get
n[5!/(2!] = 7!/(2!)
Multiply both sides by 2!; Remember 7!= 7*6*5!
Divide both sides by 5! to get
n = 7!/5!
n = 7*6
n = 42
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38. 208P(n,2) = P(16,4)
Use the definition of P(n,r) to get:
208[n!/(n-2)!] = 16!/(12!)
208[n(n-1)] = 16*15*14*13
n(n-1) = 210
n^2-n-210 = 0
(n-15)(n+14)=0
Remember, n must be positive.
n = 15
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Cheers,
Stan H.
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Cheers,
Stan H.
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