My Math Forum What is the meaning of "every choice we pick"?

 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 April 26th, 2018, 09:28 PM #1 Senior Member   Joined: Aug 2014 From: India Posts: 343 Thanks: 1 What is the meaning of "every choice we pick"? In how many ways, can you distribute 3 pencils to 8 people? In this case the order we pick people don’t matter. If I give a pencil first to A, then to B and then to C, it’s same as giving it to C first, followed by A and B. So, if we have 3 pencils to distribute, there are 3! or 6 variations for every choice we pick. What is the meaning of "every choice we pick"?
 April 27th, 2018, 05:26 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 First one way to look at this problem is that there are 8 choices to give the first pencil, then 7 people left without a pencil, so 7 choices to give the second pencil, which leaves 6 people without a pencil so 6 choices to give the third pencil. That would be $\displaystyle 8(7)(6)= 336$ ways which could also be written as $\displaystyle \frac{8(7)(6)(5)(4)(3)(2)(1)}{5(4)(3)(2)(1)}= \frac{8!}{5!}= \frac{8!}{(8- 3)!}$. But that would be much too large because it counts the same three people, in different orders, as different answers. That is wrong because "the order we pick people don’t matter" Label the people A, B, C, D, E, F, G, H. One "choice" would be to give the first pencil to "C", the second pencil to "F", and the third pencil to "D": "CFD" in that order. But since "the order we pick people don’t matter", "CDF", "DCF", "DFC", "FDC", and "FCD" are the same three people chosen in different orders and should NOT be counted as different answers. There are 3!= 6 different orders in which we could pick the same three people. "Every choice we pick", the three people we pick to give pencils to, could have been chosen in 3!= 6 different ways. All of those different ways was counted in $\displaystyle frac{8!}{5!}$. In order NOT to count those different orders we must divide by 3!. The number of way we could give 3 pencils to 8 people NOT counting different orders (who got the first pencil, who got the second) is $\displaystyle \frac{8!}{5!3!}= 56$.
 April 27th, 2018, 05:38 AM #3 Senior Member   Joined: May 2016 From: USA Posts: 1,249 Thanks: 515 They are thinking about the problem in two steps. First step is a permutation. We have 8 ways to choose the first person to get a pencil, 7 ways to choose the second person, and 6 ways to choose the third person. That is $\dfrac{8!}{(8 - 3)!} = \dfrac{8 * 7 * 6 * 5!}{5!} = 8 * 7 * 6 = 336.$ But for whatever one of those 336 choices, say we chose abc, we actually have 3! variants in terms of order of each choice of three people, namely abc, acb, bac, bca, cab, cba. But in this case we don't care about order so we go to the second step and divide our answer from the first step by 6 or 3! $56 = \dfrac{336}{6} = \dfrac{\dfrac{8!}{5!}}{3!} = \dfrac{8!}{3! * (8 - 3)!}.$ This two-step process explains the formula used for counting combinations. Last edited by JeffM1; April 27th, 2018 at 05:41 AM.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post normvcr Abstract Algebra 2 January 17th, 2013 12:06 AM lianmc Linear Algebra 3 August 10th, 2012 05:59 AM SedaKhold Calculus 0 February 13th, 2012 12:45 PM empiricus Algebra 3 June 24th, 2010 01:06 PM Niko Bellic Probability and Statistics 1 March 18th, 2010 02:15 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top