Question

Suppose a fair die is cast seven times to obtain a sequence of seven numbers. How many sequences have the property that each of the 6 possible values occurs at least once?

Answer 1

I get 6 by reasoning, but the combination formula seems to point toward 7. Any thoughts?

Answer 2

my first guess would be \(6!\times 6^2\) but i suck at these so i would not bet more than $12 it is right

Answer 3

Wow! I would bet you $20 that it is wrong! Haha

Answer 4

But thanks for trying!

Answer 5

@kropot72

Answer 6

@SithsAndGiggles

Answer 7

ok you win
but at least let me explain my thinking
you have 6 choices for the first number
5 for the second, etc giving \(6!\) ways to arrange each of the 6 different numbers
then you have 6 choices for the place for the repeat
then you have 7 spots for the repeat
ok i change my answer to
\[6!\times 6\times 7\] also probably wrong

Answer 8

I understand your reasoning. But we don't care what order they are rolled in. We just have to have them appear at least once.

Answer 9

i didn't care what order they appeared in either

Answer 10

I'm not sure if my reasoning is correct, but someone posted this question earlier to which I responded: http://openstudy.com/users/kirbykirby#/updates/54036890e4b0f2ed1e13defa

And for each choice, I was assuming you can use the same method you use when they say "How many ways can you arrange the word "MISSISSIPPI" " .. for which I said you can have \( \Large \frac{7!}{2!}\) for each of 6 choices.

Answer 11

Thanks!

Answer 12

@kirbykirby isn't that the same answer i wrote? \(7!\times 6\) ?

Answer 13

ooh i see my mistake

Answer 14

Mine sums as \[ \Large \underbrace{\frac{7!}{2!}+\ldots+\frac{7!}{2!}}_{6~of~these}=\frac{6\cdot7!}{2}\]

Answer 15

each choice is \(\frac{7!}{2}\)

Answer 16

so \(3\times 7!\)

Answer 17

ya. I hope that is right though o_O In my head it seemed logical

Answer 18

does, doesn't it

Answer 19

But all of those reasonings are based on the order being important. That 1 2 3 4 5 6 1 is a different from 2 1 3 4 5 6 1, but those are the same numbers.

Answer 20

@kirbykirby, shouldn't it be
\[ \Large \underbrace{\frac{\color{red}{6\times6!}}{2!}+\ldots+\frac{\color{red}{6\times6!}}{2!}}_{6~of~these}=\frac{\color{red}{6^2\times6!}}{2}\]
There'd be 6! positions in the sequence for the numbers we want to show up at least once, multiplied by another 6 for the 7th slot which can take on any of the six possibilities.

Answer 21

@JoannaBlackwelder I believe that they do consider the order to be important. The word "sequence" basically means "an ordered list".

@SithsAndGiggles Maybe my picture influenced your thought process, but the numbers that need to appear at least once can take up any of the 7 positions, and there is a repeating element.
I remember though there was an easy trick to remember these problems. The number of arrangements for a sequence with repeating elements is: If you have \(n_i\) symbols of type \(i\), \(i=1, 2, \ldots, k\) such that \(\sum_{i=1}^k n_i=n\), then the number of arrangements using all the symbols is \[\large \frac{n!}{\prod_{i=1}^kn_i!}\] The famous example I always saw for this was "How many ways can you arrange the word MISSISSIPPI" In which we can place the first letter in 11 ways, the second in 10 ways... etc. until we get \(11!\). But then there are 4 I's, 4 S's and 2 P's, so for any given arrangment, these can be placed in \(4!\), \(4!\) and \(2!\) ways respectively, giving: \[\frac{11!}{4!4!2!} \]

Answer 22

But aren't there only 6 possible numbers to choose from for the seventh entry?

Answer 23

If we consider 1234561,
We can put the 1st number in 7 ways:
1 _ _ _ _ _ _
1 _ _ _ _ _ _
2 _ _ _ _ _ _
3 _ _ _ _ _ _
4 _ _ _ _ _ _
5 _ _ _ _ _ _
6 _ _ _ _ _ _
Then the next number in 6 ways, etc. until we get 7!
But there is always a repeating element for a given sequence, so that's why I divided by 2. (or 2!)
Maybe cause the 7th entry is not unique to being the repeating element. Given the combination 1234561, you can also have 1213456, etc.

Answer 24

Ok well good news :) Wolfram has a feature that counts permutations. http://www.wolframalpha.com/input/?i=distinct+permutations+of+%7B1%2C+2%2C+3%2C+4%2C+5%2C+6%2C+1%7D It seems like this agrees with 7!/2

Answer 25

Ok, I see that I was looking at this incorrectly. The 7!/2! makes sense. Thanks. @kirbykirby and everyone else for your input!

Answer 26

I see too, the solution would have made more immediate sense to me if it was written as
\[\large\frac{7\times6!}{2!}=\sum_{i=1}^7\frac{6!}{2!}\]
Thanks for clearing that up!