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mathmath333
 one year ago
7 different objects must be divided among 3 people .
In how many ways can this be done if one or two
of them must get no objects ?
mathmath333
 one year ago
7 different objects must be divided among 3 people . In how many ways can this be done if one or two of them must get no objects ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{7 different objects must be divided among 3 people .}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can this be done if one or two}\hspace{.33em}\\~\\ & \normalsize \text{of them must get no objects ?} \hspace{.33em}\\~\\ & a.)\ 381 \hspace{.33em}\\~\\ & b.)\ 36 \hspace{.33em}\\~\\ & c.)\ 84 \hspace{.33em}\\~\\ & d.)\ 180 \hspace{.33em}\\~\\ \end{align}}\)

abb0t
 one year ago
Best ResponseYou've already chosen the best response.0Is this a permutations/combinations question?

Foxythepirate0984
 one year ago
Best ResponseYou've already chosen the best response.1well when you do this, everyone gets 2 objects, theres one left, so there will be an uneven amount, what do you think it is

Foxythepirate0984
 one year ago
Best ResponseYou've already chosen the best response.1dw:1443026420349:dw there is 3 people and seven objects, each one could get 2 and theres one left, there can also go 3 each, but that means 1 person will just get one object

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Consider two cases : 1) exactly two people get 0 objects 2) exactly one person gets 0 object

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3How many ways are there for Case 1 ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1a>0 b>0 c>7 a>7 b>0 c>0 a>0 b>7 c>0 total 3 way

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.32) exactly one person gets 0 object Remove one person, and consider the \(7\) different objects. Each object has \(2\) states: {person1, person2}; therefore, total possible ways for distributing these \(7\) different objects to the \(2\) people = \(2*2*2*2*2*2*2 = 2^7\) However, two of these are already counted in Case1, so total ways = \(2^72\) Again, you could remove a person in \(3\) ways, so total ways of distributing \(7\) objects to \(3\) people such that exactly one of them gets \(0\) objects is \(3(2^72)\)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i still cant understand this, \(2∗2∗2∗2∗2∗2∗2=2^7\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Remove 3rd person, so now you have only two people : {person1, person2}

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3You want to distribute the 7 objects to those two people

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Look at each object, It can go to either `person1` or `person2` so each object has `2` different states, yes ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is this equivalent to find the number of whole number soln's of a+b=7

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1yes it has 2 states

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Nope, because the 7 objects in our case are not identical.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Not just the number of objects, but also the type of object also matters

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3For example, below two are different ways : way 1 : Person1 : {o1} Person2 : {o2, o3, o4, o5, o6, o7} way 2 : Person1 : {o2} Person2 : {o1, o3, o4, o5, o6, o7}

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1looks like you need to add 3 to your solution

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is this also one of the way way 1 : Person1 : {o1,06} Person2 : {o2, o3, o4, o5, o7}

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443029763196:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ran out of room, but the pattern continues

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.1I would write \[3\cdot2^73\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I believe the way ganeshie did it, there are two cases 1) exactly two people get 0 objects : 3 2) exactly one person gets 0 object : 3(2^7 2) add them up to is there a way to do it in one case? @Zarkon

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0mathmath do you see why we subtract 2. We have to remove the two extreme cases p1,p1,p1,p1,p1,p1,p1 p2,p2,p2,p2,p2,p2,p2 that would cause there to be two people with 0 objects and that was counted in case 1,

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1Is it like 1st object can be given in 2 ways. 2nd object can be given in 2 ways. . . . 7th object can be given in 2 ways. =2^7

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1which 2 cases did u substract from 2^7

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3you can't give all the objects to just one person

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i mean there would be 2 cases within 2^7 did u substract this 2 cases as told be jaydid p1,p1,p1,p1,p1,p1,p1 p2,p2,p2,p2,p2,p2,p2

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, because they are are already accounted in Case1

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1case 1 is this right a>0 b>0 c>7 a>7 b>0 c>0 a>0 b>7 c>0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Case1 covers "all objects given to a single person"
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