My Math Forum Probability Involving Possible Outcomes of making a number.

 June 26th, 2013, 08:18 AM #1 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Probability Involving Possible Outcomes of making a number. A question was given such that $x+y+z= 8$ find the possible values of x , y , z ....given the variable MUST BE POSITIVE INTEGER. (ans : 21) FINE , i did it by listing method. from 1 + 1 + 6 , 1 + 6 + 1 , ...all the way to 3 + 2 + 2 but from the equation given , we can incur that this type of question is possible given the value must be min equal the amount of variable such that x + y + z = 3 ......it cannot be 2 because decimal takes forever to calculate for a high schooler in an exam. x + y + z = n , such that n larger or equal to the sum of total of variable. is there a way to calculate the available number of outcomes?? i don't think it is easy to do listing method because exercise question is giving me longer equation like w + x + y + z = 17 --> this is what i have done so far. assume x + y + z = 3 --> possible outcome = 1 because 3 divide by (3 variable) is 1. x + y + z = 4 --> possible outcome is 3 ........because 4/3 = 1.xx round off to 2 we get 2 + 1 + 1 --> 3C1 = 3 x + y + z = 5 .... 5/3 = 1.67 round to 2. 1 + 2 + 2 ( 3 outcome ) 1 + 1 + 3 ( 3 outcome ) total : 6 ( 3p3 = 6 ??) x + y + z = 6 ..........6/3 = 2 2 + 2 + 2 ( 1 outcome ) 4 + 1 + 1 ( 3 outcome ) 3 + 2 + 1 ( 6 outcome ) total : 10 outcome. ( 1 + 2 + 3 + 4 = 10 ??) i can't see the relationship of the progression. is there a way to calculate the available number of outcomes??
 June 26th, 2013, 07:59 PM #2 Senior Member   Joined: Sep 2012 From: British Columbia, Canada Posts: 764 Thanks: 53 Re: Probability Involving Possible Outcomes of making a numb There is, in fact, a systematic method of doing this, but only under the assumption that the integers are unordered (i.e. 2+3+6 is the same as 6+2+3). Let a+b+c=n. We can visually represent it below: * * * * * * ... where there are n "stars". We need to divide the stars into three groups (because there are three integers) with two lines, and there are n-1 gaps where are lines could be, so the number of ways to do this is just ${n-1}\choose{2}$. We could also do it with four integers: a+b+c+d=n which becomes ${n-1}\choose{3}$ In general, if there are m integers a1, a2, ... a_m that sum to n, then there are ${n-1}\choose{m-1}$ ways of choosing the values of the m integers. Note that this method only works with UNORDERED SUMS.
 June 27th, 2013, 12:03 AM #3 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Re: Probability Involving Possible Outcomes of making a numb can you use your above method to solve x + y + z = 8 for me to see? the algebra is not helping me to understand it......and i also forgot what is ${n-1}\choose{3}$ this notation means.
June 29th, 2013, 07:31 PM   #4
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Re: Probability Involving Possible Outcomes of making a numb

Hello, rnck!

Quote:
 $x\,+\,y\,+\,z \:=\:8$ Find the number of possible values of $x,\,y,\,z$ given that the variables must be positive integers. (Answer: 21)

Place eight objects in a row, leaving a space between them.
[color=beige]. . . . [/color]$\circ\, \_\,\circ\,\_\,\circ\,\_\,\circ\,\_\,\circ\,\_\,\c irc\,\_\,\circ\,\_\,\circ$

Select 2 of the 7 spaces and insert "dividers".

Then:

[color=beige]. . . [/color]$\circ\,\circ\,\circ\,\circ\,|\,\circ\,|\,\circ\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z) \:=\:(4,\,1,\,3)$

[color=beige]. . . [/color]$\circ\,|\,\circ\,\circ\,\circ\,\circ\,\circ\,|\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z) \:=\:(1,\,5,\,2)$

[color=beige]. . . [/color]$\circ\,|\,\circ\,\circ\,\circ\,|\,\circ\,\circ\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z,) \:=\:(1,\,3,\,4)$

$\text{To choose 2 objects from 7 objects, there are: }\,\frac{7!}{2!\,5!} \,=\,21\text{ ways.}$

June 30th, 2013, 11:17 AM   #5
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Re: Probability Involving Possible Outcomes of making a numb

Quote:
Originally Posted by soroban
Hello, rnck!

Quote:
 $x\,+\,y\,+\,z \:=\:8$ Find the number of possible values of $x,\,y,\,z$ given that the variables must be positive integers. (Answer: 21)

Place eight objects in a row, leaving a space between them.
[color=beige]. . . . [/color]$\circ\, \_\,\circ\,\_\,\circ\,\_\,\circ\,\_\,\circ\,\_\,\c irc\,\_\,\circ\,\_\,\circ$

Select 2 of the 7 spaces and insert "dividers".

Then:

[color=beige]. . . [/color]$\circ\,\circ\,\circ\,\circ\,|\,\circ\,|\,\circ\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z) \:=\4,\,1,\,3)" />

[color=beige]. . . [/color]$\circ\,|\,\circ\,\circ\,\circ\,\circ\,\circ\,|\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z) \:=\1,\,5,\,2)" />

[color=beige]. . . [/color]$\circ\,|\,\circ\,\circ\,\circ\,|\,\circ\,\circ\,\c irc\,\circ\;\text{ represents }\,(x,\,y,\,z,) \:=\1,\,3,\,4)" />

$\text{To choose 2 objects from 7 objects, there are: }\,\frac{7!}{2!\,5!} \,=\,21\text{ ways.}$

I recall you showing this once before and, as I did before, I just have to marvel at the brilliance of this visual demo.

 July 1st, 2013, 05:11 AM #6 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Re: Probability Involving Possible Outcomes of making a numb wow thats insane....how u decide where to put divider? is n(divider) = n(variable)-1 ? and after u put it , how come suddenly pop choosing 2 objects from 7 objects? 7! seems understandable but the denominator 2!5! come from where? you gotta tell me man! my whole school even my teach cant formulate out this piece of math!!
 July 1st, 2013, 07:21 AM #7 Senior Member   Joined: Apr 2010 Posts: 128 Thanks: 0 Re: Probability Involving Possible Outcomes of making a numb w + x + y + z = 17 if there is 16 gap , and 3 divider... $\frac{16!}{3!.(16-3)!}=560= \ ^{16}C_{3}$ indeed it is answer..... -Edit : ......i had understand how it works. can close topic le. Indeed , your illustration is MARVELOUS.....it looks so familiar i suddenly realised i've been tricked. I never would have thought that a simple question of asking how to arrange 2 red chair in between 8 green chair such that red chair must be placed in between 2 green chair can be asked in this way........

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