Ryan reaches into his gym bag for a pair of running shoes. If there are seven different pairs of shoes loose in the bag, how many ways can he pick two shoes that do not match?
what I did is
Total= 14*13=182
Matched pairs=7*2=14
182-14- 168 possibilities
combinations of 14 things two at a time = 14!/[12!(2!)]=14*13/2 = 7*13 = 91
but some of those are matching, namely 7 pairs, so there are 91-7 = 84 ways
Why did you divide the answer from 14*13 by 2?
23! = 2*1 = 2
excuse typo
2! = 2*1 = 2
combinations is
n!/[ (n-r)! r! ]
here n=14
r = 2
14! = 14*13*12 * 11 etc
12! =12*11 etc
so
14!/12! = 14 * 13
now divide that by 2! which is 2
but you can not tell left-right
from
right-left
well for the matched pairs part the 7 is the pairs and the 2 is the orders of pairs LR and RL
To find out how many ways Ryan can pick two shoes that do not match, we need to determine the number of ways he can choose two shoes from the available seven pairs.
To begin, we define the shoes in each pair as distinct entities. This means that each shoe in a pair is different from the other shoe, and we can differentiate between them.
Since there are seven different pairs of shoes in the bag, and each pair consists of two distinct shoes, there are a total of 7 * 2 = 14 different shoes in the bag.
Now, to count the number of ways Ryan can choose two shoes that do not match, we can utilize the concept of combinations. Since we want to choose two shoes from the available 14 shoes, without any restriction on the order, we can calculate the number of combinations using the formula:
C(n, r) = n! / (r! * (n-r)!)
where C(n, r) is the number of combinations of choosing r items from a total of n items, and "!" denotes the factorial of a number.
Plugging in the values, we have:
C(14, 2) = 14! / (2! * (14-2)!)
= 14! / (2! * 12!)
= (14 * 13) / (2 * 1)
= 91
Hence, there are 91 ways Ryan can pick two shoes that do not match from the seven different pairs in his gym bag.