Determine the value of the following (a) 4P2 (b) 9P3 (c) 20P2
Answers
Explanation:
1) a) 4P2 = 4!/(4-2)! = 4!/2! = (4*3*2!)/2! = 4*3 = 12 EXCEL function: PERMUT(4, 2) b) 7P5 =7!/(7-5)! = 7!/2! = (7*6*5*4*3*2!)/2! = 7*6*5*4*3 = 2520 EXCEL function: PERMUT(7, 5) c) 10P3 = 10!/(10-3)! = 10!/7! = (10*9*8*7!)/7! = 10*9*8 = 720 EXCEL function: PERMUT(10, 3) 2) (n+1)P3 = (n+1)!/(n+1-3)! = (n+1)!/(n-2)! = [(n+1)(n)(n-1)(n-2)!]/(n-2)! = (n+1)(n)(n-1) nP4 = n!/(n-4)! = [(n)(n-1)(n-2)(n-3)(n-4)!]/(n-4)! = (n)(n-1)(n-2)(n-3) So to find n, we have to equate the above two values and solve for n: (n+1)(n)(n-1) = (n)(n-1)(n-2)(n-3) Cancelling the common terms (namely n and n-1) from both sides, we get; (n+1) = (n-2)(n-3) n+1 = n2 – 3n – 2n +6 0 = n2 – 6n + 5 Solving through the quadratic equation n=6±(√−6)2−4∗1∗52∗1 which gives us n = 5, 1, and since we're considering permutations, we only take n = 5 3) Number of ways 5 people can be seated on a sofa if there are only 3 seats available: 5P3 = 5!/(5-3)! = 5!/2! = (5*4*3*2!)/2! = 5*4*3 = 60 ways 4) a) Number of ways 7 books can be arranged on a shelf if any arrangement is possible: Suppose these are the seven possible places that the books can take _ _ _ _ _ _ _ .
p=2
Answer:
p=3
p=10
Explanation:
it's the answer of value p