Combinations problem

[dbag]

Hello

Problem statement: 13 boxes can each have one thing out of 3 options (1, 2 or a circle). How many different arrangements are there, where 10 of these boxes have a "1" and the rest (3) can have anything else?

Solution: 13!/(10!)*3!=286

Is that right?

 

[pka]

Problem statement: 13 boxes can each have one thing out of 3 options (1, 2 or a circle). How many different arrangements are there, where 10 of these boxes have a "1" and the rest (3) can have anything else?
Solution: 13!/(10!)*3!=286
Is that right?

That counts the number of arrangement of the string \(\displaystyle 1111111111000\).
What about the strings \(\displaystyle 1111111111222\) or \(\displaystyle 1111111111220\) or \(\displaystyle 1111111111200\) ?

 

[dbag]

the last 3 could be either 2 or o. so yes i should take them into account aswell

13!/10!*2*3! =143 or rather it adds more possibilities so divide the 3! at the end and get 572?

 

[pka]

the last 3 could be either 2 or o. so yes i should take them into account aswell
13!/10!*2*3! =143

I will give you a back-of-the-book answer and let you explain why.
\(\displaystyle 2\cdot\frac{13!}{(10!)(3!)}+2\cdot\frac{13!}{(10!)(2!)}\)

 

[dbag]

the first part takes into account the 000's and the second parts takes into account the 222's

 

[pka]

the first part takes into account the 000's and the second parts takes into account the 222's

No, first counts \(\displaystyle 1111111111000\text{ or }1111111111222\) the second counts \(\displaystyle 1111111111002\text{ or }1111111111220\)

 

[dbag]

i see. So the 3! is the amount of same type, and the 2! is 2 0's or 2 2's. So if we had 4 options would the problem change to 3*(13!/10!*4!)+3*(13!/10!*3!) ?

 

[Steven G]

the last 3 could be either 2 or o. so yes i should take them into account aswell

13!/10!*2*3! =143 or rather it adds more possibilities so divide the 3! at the end and get 572?

Please use parenthesis. 13!/10!*2*3! means \(\displaystyle \dfrac{13!}{10!}*2*3!\)

 

[Steven G]

Here is how I would think about it. There are ₁₃C₁₀ways to choose the 10-1s. Then for the last three positions there are 2^3 ways of doing that (2 choices for each of the three).

So my final answer will look like ₁₃C₁₀*2^3

 

[Steven G]

the first part takes into account the 000's and the second parts takes into account the 222's

If you post is correct then pka must be wrong since it (based on your post) does not take into account when you have both 0s and 2s. Do you see that? For the record pka is usually never wrong.

 

[pka]

i see. So the 3! is the amount of same type, and the 2! is 2 0's or 2 2's. So if we had 4 options would the problem change to 3*(13!/10!*4!)+3*(13!/10!*3!) ?

In sections on counting this is often called the \(\displaystyle MISSISSIPPI\) question.
There are eleven letters there. But there are fewer than \(\displaystyle 11!\) ways to rearrange that word because some letter are indistinguishable.
If it were \(\displaystyle MI_1S_1S_2I_2S_3S_4I_3P_1P_2I_4\) we now have eleven different letters and \(\displaystyle 11!\) ways to rearrange.
Lets remove the subscripts from the I's \(\displaystyle MI_1SS_2IS_3S_4IP_1P_2I_4\)
This is the key the string \(\displaystyle I_1 I_2 I_3 I_4\) can be arranged in \(\displaystyle 4!=24\) ways. So we must divide them off.
As the number of ways to rearrange \(\displaystyle MISSISSIPPI\) is \(\displaystyle \dfrac{11!}{(4!)^2(2!)}\)

 

[dbag]

If you post is correct then pka must be wrong since it (based on your post) does not take into account when you have both 0s and 2s. Do you see that? For the record pka is usually never wrong.

my post was wrong as pka stated. first counts 1111111111000 or 11111111112221111111111000 or 1111111111222 the second counts 1111111111002 or 1111111111220. i understand your solution, how is it different from what pka said? after all does it not take into account the last 3 possibilities from 2 options? 2-2's or 2-0's

 

[Steven G]

my post was wrong as pka stated. first counts 1111111111000 or 11111111112221111111111000 or 1111111111222 the second counts 1111111111002 or 1111111111220. i understand your solution, how is it different from what pka said? after all does it not take into account the last 3 possibilities from 2 options? 2-2's or 2-0's

The two results are identical.

For the last three: You have 2 choices for each, giving 8 possibilities. They are 000, 002, 020, 200, 220, 202, 022 and 222. What did I not include?

 

[dbag]

you forgot nothing, i am the only one here who seems to be confusing things. Thanks to both of you

 

[pka]

The two results are identical.
For the last three: You have 2 choices for each, giving 8 possibilities. They are 000, 002, 020, 200, 220, 202, 022 and 222. What did I not include?

Content wise, that are only four possibles: \(\displaystyle 000\;002\;022\;222\)
Consider the strings \(\displaystyle 1111111111000\text{ or }1111111111222\) there are \(\displaystyle \dfrac{13!}{10!\cdot 3!}\) ways to rearrange each string.
Now consider the strings \(\displaystyle 1111111111002\text{ or }1111111111022\) there are \(\displaystyle \dfrac{13!}{10!\cdot 2!}\) ways to rearrange each string.

If there were a third possibility, say Consider the strings \(\displaystyle 0,2,3\) .
Then content wise there at ten possibilities: \(\displaystyle 000\;002\;003\;022\;023\;033\;222\;223\;233\;333\) that is the number of ways to put three identical objects (in this case choices) into three different cells \(\displaystyle \mathcal{C}^5_2=\dbinom{5}{2}\)