Question 755961: Hello! :)
1. How many 4 digit numbers greater than 1000 can be formed with no repetition in digits?
2. Suppose you want to hang 5 different pairs of socks on a clothesline to dry. How many ways can you arrange the socks on the line?
Thanks so much for the help I really appreciate it!!
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you want 4 digit number.
first digit can be any of 1 to 9 so that gives you 9 choices. second digit can be any of 0 to 9 so that gives you 10 choices. third digit can be any of 0 to 9 so that gives you 10 choices. last digit can be any of 1 to 9 so that gives you 9 choices.
the number of possible numbers would be 9 * 10 * 10 * 9 = 8100 ways.
the number will be any of 1001 to 9999.
you have 5 different pairs of socks.
that's 10 socks in total.
assuming each sock is to be treated as different from each other sock, you can arrange them 10 factorial ways.
that's 10! which is equal to 10*9*8*7*6*5*4*3*2*1 = 3628800 ways.
assuming each sock in a pair is to be treated as identical to its twin, but different from any other sock in any other pair, then the number of possible arrangements would be equal to 10! / (2!*2!*2!*2!*2!) ways which is equal to 113400 ways.
the basic formula for arranging is n! / x! where n is the total number of elements in the set and x is the number of elements that are identical to each other.
an easy way to see this is the letters a,b,c.
these can be arranged in 3! = 6 ways as follows:
abc
acb
bac
bca
cab
cba
assume the letters are a,b,b instead. these can be arranged in 3!/2! = 3 ways as follows:
abb
bab
bba
no other unique arrangements are possible because 2 of the letters are the same.
in the formula, this was taken care of by dividing by 2!. that is the total number of ways the first b and the second b could have been arranged if they ere different.
your socks problem is difficult because it could be interpreted in different ways. not being sure what your instructor is expecting, i gave you some options. if it's a simple treat each sock as unique problem, then 10! would be the answer. if you are to treat each pair of socks as being identical then 10! / (2!*2!*2!*2!*2!) would be the answer. if all the socks were considered identical than only 1 arrangement would be possible.
one other possibility comes to mind. if each pair of socks was hung as a unit, then you would have 5! ways to hang them because there would only be 5 slots on the line rather than 10 slots.
sorry for the confusion but there's too much ambiguity in the problem statement to definitively say one answer is preferable to the other. you would need more information to detail exactly what they had in mind.
the number problem, on the other hand, was clear with no ambiguity and so was much easier to solve.
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