Permutations homework question

[six789]

i just want to ask if my answer is correct...
the problems is ... how many numbers greater than 300 000 are there using only the digits 1,1,1,2,2,3?

my answer is... 1(three)*5(the position)*2(the numbers 1 and 2) = 10
therefore, the numbers greater than 300 000 is 10

is this right? thanks for the replies...

 

[NateTG]

Hmm...
111223
111232
111322
113122
131122
311122
311221
311212
312112
321112
321121

Clearly there are more than 10.

 

[six789]

Hmm...
111223
111232
111322
113122
131122
311122
311221
311212
312112
321112
321121
Clearly there are more than 10.

i think you are misunderstanding the question... the ways to get combinations greater than 300,000

Hmm...
311122
311221
311212
312112
321112
321121

this ones are right... but I am asking if it is ten?

 

[Erzeon]

its 1 * 5! , I think at least. the 3 has to be in front, and then you got 5 more numbers that can go anywhere. so 5! is 120. Therefore 120 numbers greater than 300,000.

 

[six789]

its 1 * 5! , I think at least. the 3 has to be in front, and then you got 5 more numbers that can go anywhere. so 5! is 120. Therefore 120 numbers greater than 300,000.

i don't think so cause you have 3 ones and 2 twos... i know what you mean, but the number is not different and that is applicable only if the numbers is different

 

[Erzeon]

Oh ok, my bad. Let me think:D

 

[Erzeon]

5!/(3! x 2!)

is the answer I am pretty sure. which is 10 which means you're right.

 

[six789]

5!/(3! x 2!)
is the answer I am pretty sure. which is 10 which means you're right.

for the first time, lol, I am right, *claps*

 

[HallsofIvy]

There are, of course, an infinite number of numbers "larger than 300000" with any digits at all!
You seem to be assuming "larger than 300000 but less than 400000". If that was the problem, then it would help a lot for you to say so!

 

[six789]

There are, of course, an infinite number of numbers "larger than 300000" with any digits at all!
You seem to be assuming "larger than 300000 but less than 400000". If that was the problem, then it would help a lot for you to say so!

ok then thanks for the info

 

[HallsofIvy]

Now that I read it again, I see that I misunderstoond the question. I was thinking you meant 'numbers larger than 300000 using the digits "1", "2", and "3"' but since you are repeating the "1", you clearly mean just using those 6 digits.
To be larger than 300000, the first digit must be 3 so it's really "in how many ways can we write the 5 digits 1, 1, 1, 2, 2". That's really a "binomial" problem: it is [itex]\frac{5!}{3!2!}= 10[/itex]. One way of seeing that is to imagine that each of those had a subscript: 1₁, 1₂, 1₃, 2₁, 2₂ so that they are distinguishable. There are 5! ways of ordering those 5 symbols. Now we not that 3! ways of rearranging just the 3 "1"s so that for each of the ways of rearranging those 5 "distinguishable" symbols there are 5 more that are just rearranging the different "1"s. Since they are NOT distinguishable, we don't want to count those as 6 different ways so we divide by 3!= 6. The same thing happens with the "2"s. In order NOT to count as different orders that just swap the two "2"s, we divide by 2!= 2.

 

[six789]

Now that I read it again, I see that I misunderstoond the question. I was thinking you meant 'numbers larger than 300000 using the digits "1", "2", and "3"' but since you are repeating the "1", you clearly mean just using those 6 digits.
To be larger than 300000, the first digit must be 3 so it's really "in how many ways can we write the 5 digits 1, 1, 1, 2, 2". That's really a "binomial" problem: it is [itex]\frac{5!}{3!2!}= 10[/itex]. One way of seeing that is to imagine that each of those had a subscript: 1₁, 1₂, 1₃, 2₁, 2₂ so that they are distinguishable. There are 5! ways of ordering those 5 symbols. Now we not that 3! ways of rearranging just the 3 "1"s so that for each of the ways of rearranging those 5 "distinguishable" symbols there are 5 more that are just rearranging the different "1"s. Since they are NOT distinguishable, we don't want to count those as 6 different ways so we divide by 3!= 6. The same thing happens with the "2"s. In order NOT to count as different orders that just swap the two "2"s, we divide by 2!= 2.

thanks again for the info... it help a lot on my understanding regarding permutations... can you check my other post, which is "probability".