Question

How many different seven-digit numbers can be produced by rearranging the digits of 2502254 ?

Answer 1

you have seven digits there, and so if they were all different, the answer would be \(7!\)

Answer 2

but since you cannot tell one "2" from another, you have to divide that by \(3!\)
your answer is therefore
\[\frac{7!}{3!}\]

Answer 3

but how do you know that none of those numbers are repeats?

Answer 4

i thought you have to do it by making 0,2,4,5 each leading numbers and going from there

Answer 5

since it said "rearranging" that is what i assumed

Answer 6

oh damn my answer was wrong anyway!

Answer 7

i think the answer is 360...i looked it up but it didn't make sense to me.

Answer 8

you have two fives as well, so it is
\[\frac{7!}{3!2!}\] do you understand this notation ?

Answer 9

i get 420

Answer 10

kind of....why is it only over 3! and 2! though?

Answer 11

ok lets imagine we had 7 completely different number, say the digits
1,2,3,4,5,6,7
and we wanted to know how many ways we could arrange them.
7 choices for the first digit, then
6 for the second (since we used up one) then
5 for the third, then
4 then
3 then
2 then 1

by the "counting principle" the number of ways we can do these things all together is
\[7\times 6\times 5\times 4\times 3\times 2\times 1 =7!\]

Answer 12

so if they were all different, the number of ways to arrange them would be \(7!\) which is just shorthand for what i wrote above

Answer 13

http://mvtrinh.wordpress.com/ i found this online and i have no idea what he did. i understand the way you're expaining it but i'm not sure which is right

Answer 14

well this looks more complicated that what i wrote, but let me read it for a second

Answer 15

okay thanks so much!

Answer 16

aah ok lets continue with the method i wrote.
the answer here is correct, but i think needlessly complicated

Answer 17

going back to the counting principle, we see if they are all distinct there would be \(7!\) different arrangements.
however, since we cannot tell the three "2"s apart, we have counted by \(3!=6\) too many ways. is that clear?

Answer 18

that is the number of ways you can "permute" the three "2"s

Answer 19

yeah i got that

Answer 20

similarly, we cannot tell the two "5"s apart, so we have overcounted also by a factor of \(2!=2\)

Answer 21

therefore i thought the answer was
\[\frac{7!}{3!2!}=420\] but i made a MISTAKE!
i forgot that you cannot begin a 7 digit number with zero!!

Answer 22

ohhh! okay i get the zero part..but how would you account for that?

Answer 23

so we subtract the ones we counted with a zero at the beginning

Answer 24

how many are there? by the exact same reasoning, now that i have put zero in the first spot, there would be \(\frac{6!}{3!2!}=60\) possibilities, therefore we have over counted by 60

Answer 25

and low an behold \(420-60=360\) just like it says in the book
i totally ignored that zero business, but now we both learned something

Answer 26

good catch by the way, finding that link
on the other hand i think it is more complicated that necessary. on the other hand, as usual with math, there is
a) more than one way to skin a cat and
b) different interpretations depending on your point of view

Answer 27

haha ! thank you soooooo much! you're a life saver!

Answer 28

glad to help
do you have to write this up and turn it in, or just get the answer, or none of the above (just curious)?

Answer 29

write it up and turn it in

Answer 30

ok, then don't lose this thread! should be clear now though, have fun

Answer 31

i won't haah thanks again!

Answer 32

yw (again)