
Advanced Statistics Advanced Probability and Statistics Math Forum 
 LinkBack  Thread Tools  Display Modes 
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 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 n1 gaps where are lines could be, so the number of ways to do this is just . We could also do it with four integers: a+b+c+d=n which becomes In general, if there are m integers a1, a2, ... a_m that sum to n, then there are 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 this notation means. 
June 29th, 2013, 07:31 PM  #4  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Probability Involving Possible Outcomes of making a numb Hello, rnck! Quote:
Place eight objects in a row, leaving a space between them. [color=beige]. . . . [/color] Select 2 of the 7 spaces and insert "dividers". Then: [color=beige]. . . [/color] [color=beige]. . . [/color] [color=beige]. . . [/color]  
June 30th, 2013, 11:17 AM  #5  
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: Probability Involving Possible Outcomes of making a numb Quote:
 
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... 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........ 

Tags 
involving, making, number, outcomes, probability 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Problems involving number bases  Michaeld71  Algebra  3  November 25th, 2012 04:17 PM 
Number of possible outcomes when driving  groundwar  Algebra  6  October 27th, 2009 02:08 PM 
probability of making a straight?  durexlw  Algebra  1  February 9th, 2009 08:44 PM 
Probability: making at least a double pair out of 52 cards  durexlw  Algebra  4  February 5th, 2009 02:40 PM 
a probability/gambling/odds making question  captainglyde  Algebra  1  November 30th, 2007 11:16 AM 