SOLUTION: what is the answer here
nP2 = 380 *
10P5 = ? *
8Pr = 20160 *
6P6 = ? *
7Pr = 840 *
nP3 = 60 *
12P3 = ? *
nP4 = 7920 *
i needed before 11 am here in
Solving these problems by going back to the formal definition of nPr, as the other tutor did, is absurd.
Use the common sense definition of nPr -- it is the product of r consecutive integers, starting with n and counting down.
Let's look at these problems in groups of similar problems....
(1) 10P5, 6P6, and 12P3.
These are straightforward calculations.
10P5: product of 5 consecutive integers starting with 10 and counting down: 10*9*8*7*6
6P6: 6*5*4*3*2*1 (= 6!; note that 6P5 = 6*5*4*3*2 is also equal to 6!)
12P3: 12*11*10
You can do the calculations.
(2) 8Pr=20160; 7Pr=840
Simply start with n and multiply by decreasing integers until you get the desired result.
8Pr=20160: 8*7=56; 56*6=336; 336*5 = 1680; 1680*4 = 6720; 6720*3 = 20160 ANSWER: r=6 (remember, the answer is not the last number you multiplied by; the answer is the number of integers you multiplied together to get the 20160)
Do the 7Pr=840 problem the same way.
(3) nP2=380; nP3=60; nP4=7920
These are the hardest. You can solve them by trial and error; but here is way to home in on the answers for this kind of problem quickly.
nP2=380....
nP2 is the product of two consecutive integers, so it is close to a number squared. 380 is close to 400, which is 20 squared, so try 20*19.
20*19 = 380; it works.
ANSWER n=20
nP3=60....
nP3 is the product of three consecutive integers, so it is close to a number cubed. 60 is close to 64, which is 4 cubed. So try 5*4*3.
5*4*3=60; it works.
ANSWER: n=5
nP4=7920
mP4 is the product of FOUR consecutive integers, so it is close to a number to the 4th power. 10^4 is 10,000; 9^4 is 6561. So try 10*9*8*7.
10*9*8*7 = 5040; too small. so try 11*10*9*8.
11*10*9*8 = 7920; it works.
ANSWER: n=11.