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mathmath333
 one year ago
Counting Problem
mathmath333
 one year ago
Counting Problem

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{In how many ways a 12 member team can be divided }\hspace{.33em}\\~\\ & \normalsize \text{into 3 groups such that each group has 4 members.}\hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0try this for the first group you want to choose 4, and there are 12 to choose from 12 choose 4 second group, there are 8 left to choose from, and you want to choose 4 again 8 choose 4 there are 4 left , and you need to choose 4 again 4 choose 4 by multiplication principle 12 choose 4 * 8 choose 4 * 4 choose 4

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0i tried this it says wrong

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.212 people and 3 groups of 4 no. of ways to select 4 people from 12 for first group = 12C4 = 495 no. of ways to select 4 people from remaining 8 for second group = 8C4 = 70 no. of ways to select 4 people from 4 for third group = 4C4 = 1 total no. of ways to select people for group = 495 * 70 * 1 / 3! = 34650 /6 = 5775 Here we divide by 3! here position or identification of groups does not matter. All groups have equal identity, arrangement of group does not matter. For example, if there are 3 groups having names G1, G2, G3 then there is no need to divide by 3!. Sometimes there may be typing mistake in solution of a problem, so please ignore it. Understand the concept and try the problem yourself.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2u can go here  http://www.algebra.com/algebra/homework/Permutations/Permutations.faq.question.342414.html :P

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0actually there are 2 answers given i m not sure which is correct. \(\large \color{black}{\begin{align} & \binom{12}{3}\times \binom{9}{4}\hspace{.33em}\\~\\ \end{align}}\) \(\large \color{black}{\begin{align} & \dfrac{12!}{4!*4!*4!*3!}\hspace{.33em}\\~\\ \end{align}}\) Book has errors in the past

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0thanks to imquerty.

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2:D ur welcome :P i'll take out a nice name for u :) :P

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but why he divided by 3! lol

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.2we took the product and divided by 3! since we don't care about the order of the groups :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lets say we label the players from 1 to 12 Lets look at one particular partition into three teams : {1,2,3,4} {5,6,7,8} {9,10,11,12} {1,2,3,4} {5,6,7,8} {9,10,11,12} = {1,2,3,4}{9,10,11,12} {5,6,7,8} ={5,6,7,8} {1,2,3,4} {9,10,11,12} ={5,6,7,8} {9,10,11,12} {1,2,3,4} = {9,10,11,12}{1,2,3,4} {5,6,7,8} = {9,10,11,12} {5,6,7,8} {1,2,3,4} These 6 partitions are equivalent (indistinguishable). It doesn't matter if players {1,2,3,4} are in group 1, or in group 3, since all we care is that the players are subdivided into three groups. This is true in general for each partition, so we are overcounting per partition by a factor of 3! or 6. To fix this we divide by 3!
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