how many different ways can eight cars finish in a race 1st, 2nd, 3rd ?
72
336
Ok if you have cars A B C D E F G H. Your options for the first three places are :
ABC, ABD, ABE, ABF, ABG, ABH
ACB, ACD, ACE, ACF, ACG, ACH
ADB, ADC, ADE, ADF, ADG, ADH
AEB, AEC, AED, AEF, AEG, AEH
AFB, AFC, AFD, AFE, AFG, AFH
AGB, AGC, AGD, AGE, AGF, AGH
AHB, AHC, AHD, AHE, AHF, AHG
6 x 7 = 42 options with car A in first place.
you can go through and do the same thing with each car in first place...
BAC, BAD, BAE..... HGD, HGE, HGF...
But to do it mathematically ~ 8 cars, 42 options with each car in first place.
42 x 8 = 336 different ways eight cars can finish a race in first, second and third place.
1,000,000,000
To determine the number of different ways the eight cars can finish in the race, you need to use the concept of permutations.
The first place in the race can be taken by any one of the eight cars. Once the first place has been decided, there are seven remaining cars that can take the second place. Finally, once the first and second places have been determined, there are six cars left to take the third place.
Therefore, the number of different ways the eight cars can finish in the race 1st, 2nd, 3rd is calculated as:
8 x 7 x 6 = 336 different ways.
So, there are 336 different ways the eight cars can finish in the race.