Question

How many numbers greater than 6000000 can be formed with 1, 1, 6, 6, 9, 9, 0? (repetitions not allowed)

Answer 1

by no repetition I'm guessing you mean we must use every listed digit the number of times it appears -- we're only free to permute

Answer 2

9966110,6961091,9669110,6699110, and a lot more

Answer 3

2880

Answer 4

in that case... recognize that we sure as hell can't have 1xxxxxx and that's pretty much the only restriction -- every other combination is sure to be MUCH nicer.

recognize there are \(7!\) ways to permute your list but due to duplicates we divide by \(2!\) (for the two ones) then \(2!\) again (for the two sixes) and lastly \(2!\) again (for the two nines) giving us \(7!/(2!2!2!)=7/2^3=7!/8\) possible 7-digit numbers

Answer 5

now we will systematically exclude those that start with 1

Answer 6

and 0

Answer 7

lol

Answer 8

\(1xxxxxx\) leaves \(6\) digits free which may be filled in \(6!\) ways but only \(6!/(2!2!)=6!/2^2=6!/4\) *distinct* numbers can be formed this way. \(0xxxxxx\) provides a similar result but \(6!/(2!2!2!)=6!/2^3=6!/8\)

our answer is then just \(7!/8-6!/4-6!/8=6!/8\cdot(7-2-1)=360\)

let me know if this is incorrect and you spot any errors

Answer 9

but i coudnt understand much @oldrin.bataku

Answer 10

@dg2 notice that all numbers beginning with 6 or 9 are guaranteed to be larger than 6000000 but those beginning with 0 or 1 are for sure smaller

Answer 11

u r right but cudnt

Answer 12

right so 1 cannot be the first number i know it very well

Answer 13

it can be only 6 and 9 only 2!

Answer 14

you can do it without inclusion-exclusion you just get \(6!/2^2+6!/2^2=6!/2=360\) but I originally forgot about the \(0\) and already began to do it my inclusion-exclusion way so I just stuck with it

Answer 15

the way i did it was by counting the total number of possible numbers formed by the digits which is \(7!/(2!2!2!)\) -- this is because there are \(7!\) ways to permute the numbers and for each of the 3 digits that appears twice, there are \(2!\) permutations of those identical digits that preserve the number (i.e. imagine swapping the 3s in 1337 -- the number is still 1337 :-)

Answer 16

then I counted the number of numbers that began with \(1\) leaving \(6\) digits left undetermined to be filled with the remaining \(0,1,6,6,9,9\); like above there are \(6!\) raw permutations of the digits but due to the \(2\) digits that appear twice we divide by \(2!2!\) to only count distinct numbers i.e. \(6!/(2!2!)\)

Answer 17

I did the same process for counting the numbers that began with \(0\) only this time there are now \(3\) digits that appear twice (i.e. \(1\),\(6\),\(9\)) so we divide by \(2!2!2!\) and get \(6!/(2!2!2!)\)

Answer 18

after counting the number of numbers I want to exclude (i.e. how many begin with \(0\) or \(1\)) I subtract from the total count \(7!/(2!2!2!)\) to leave behind only those that do NOT begin with \(0\) or \(1\) giving us \(7!/(2!2!2!)-6/(2!2!)-6/(2!2!2!)=6!/2=360\)

Answer 19

wow great!.but it takes little bit time to understand ur concept

Answer 20

you can also count the numbers that begin with \(6\) or \(9\) directly which gives you \(6!/(2!2!)\) both times resulting in \(6!/(2!2!)+6!/(2!2!)=6!/2=360\) such numbers

Answer 21

First number can't be 6 and 0. then just use the choose function.

Answer 22

I'll readily admit inclusion-exclusion was overkill for this particular problem BUT it worked and it's a valuable technique to gain experience with

Answer 23

you can also do it the way whoever was here earlier did it i.e. \(4\cdot6\cdot5\cdot4\cdot3\cdot2\) but make sure to recognize this does not account for duplicates; for each duplicated digit you must divide by \(2!\) to count *distinct* numbers so \((4\cdot6\cdot5\cdot4\cdot3\cdot2)/(2!2!2!)=360\)

Answer 24

4C1 * 5C1 * 4C1 * 3C1 *2C1 *1

Answer 25

unfortunately that is incorrect mebs :-/