probability

[astral glow]

In how many ways can 11 # signs and 8* signs be arranded in a row so that no two * signs come together?

[dsaqwert]

wrong solution...

[arjmen]

Vote 12C8 = 495

[ish]

Vote 12C8 = 495

I agree with Arj. You've solved this before in the PS forum havent you?

[manwiththemission2005]

Ish, Can you explain the answer ? :-)

[arjmen]

I agree with Arj. You've solved this before in the PS forum havent you?

Yep.

 

Here:

http://www.www.urch.com/forums/showthread.php?t=29195

[manwiththemission2005]

Thanks arjmen. You have saved Ish's time . LOL . Thanks again

[ish]

You have saved Ish's time .

:D :D ...

[astral glow]

ok..ppl..

arjmen., i read ur thread

me still got a doubt

shouldnt it be 12P8

theres afterall, difference betwen the 1st* and the 2nd*...rite?

confused..help

[manwiththemission2005]

* of the same type

# of the same type. They didnt specify it as different too

[astral glow]

ok... i still dont ustand

then there is still is a difference betn

1* 2* 3* 4*

and

4* 3* 2* 1*

[1*-is the first asterisk n so on..)rite?

then how can combn be used?

[ish]

right. there is a difference between 1.2.3.4 and 4.3.2.1 (in terms of order, and not value) because the items are clearly distinct.

But not so with arranging **** or ####.

Do you get it now?

You must have tackled sums where you have to find the different words that can be made from words like "mathematics" and so on. You divide by the number of letters that repeat, dont you? This is essentially what we are doing here, in that I arrange 11 # signs in 11! ways, but since they are identical, I divide by 11! to account for that. So in the end all you need to calculate is 12C8

[manwiththemission2005]

Yep. Nice explanation Ish

[Dimas]

Another solution -- the way I looked at the question.

 

I built this initial block:

 

* # * # * # * # * # * # * # *

 

Now we have four # to distribute at any place before, after, or within the previous block. The avilable locations are numerated.

 

Then the problem is: how to find the different combinations of four items distributed to nine locations. Or, how to choose four numbers from 1 to 9 with repetition without order. The solution is (9+4-1)C4 = 12C4. Note that I used the general formula:

 

mCCn = (m+n-1)Cn. Where CC is the combinations that allows repetition.

 

Another solution is using Stirling problem (fourth type). The question reads as:

How many number of combinations you can distribute 11 identical items into 9 distinct locations where each location is at least occupied by one #. Let's say the solution is A. But wait, it is not incumbent to place # at the first and last locations. So, we should calculate how to distribute 11 items to 8 distinct locations, which represent the possibility of placing at least a single # on location 1 beside locations from 2 to 8. Let's say the output is B. But wait, we shoul multiply it by 2 because we have location 8 as well. But wait, we can place them in only the interior 7 locations which willl give us C combinations.

 

The solution is S4(11,9)+2*S4(11,8)+S4(11,7)

 

I don't have the S4 table or formulas. But this is my conjecture. It is interesting to hear your comment Arjmen.

 

By the way, the Stirling solution is way much more complicated that Arjmen's or mine. I wanted to provide it because it gives an insight for Partitioning problems which you may encounter in another problems. Good understanding of Partitioning is invaluable.

[astral glow]

thanks a lot guys..it helps

[SangFroid]

Arjmen,

 

first of all, you always ace here. your logics sound too nice...:tup:

 

and I also appreciate Holden_Caulfield :tup:

 

But i am confused with this one... why are u taking combination over here ?? can u pls elaborate it a bit more ?

 

 

Yep.

 

Here:

http://www.www.urch.com/forums/showthread.php?t=29195

[rishanthchavali]

Yep.

 

Here:

http://www.www.urch.com/forums/gmat-problem-solving/29195-number-ways.html

 

You are genius boss!!