Question 1178388
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Solving these problems by going back to the formal definition of nPr, as the other tutor did, is absurd.<br>
Use the common sense definition of nPr -- it is the product of r consecutive integers, starting with n and counting down.<br>
Let's look at these problems in groups of similar problems....<br>
(1) 10P5, 6P6, and 12P3.<br>
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.<br>
(2) 8Pr=20160; 7Pr=840<br>
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)<br>
Do the 7Pr=840 problem the same way.<br>
(3) nP2=380; nP3=60; nP4=7920<br>
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.<br>
nP2=380....<br>
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<br>
nP3=60....<br>
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<br>
nP4=7920<br>
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.<br>