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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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 ?

Mathematics
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\(\large \color{black}{\begin{align} & \normalsize \text{7 different objects must be divided among 3 people .}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can this be done if one or two}\hspace{.33em}\\~\\ & \normalsize \text{of them must get no objects ?} \hspace{.33em}\\~\\ & a.)\ 381 \hspace{.33em}\\~\\ & b.)\ 36 \hspace{.33em}\\~\\ & c.)\ 84 \hspace{.33em}\\~\\ & d.)\ 180 \hspace{.33em}\\~\\ \end{align}}\)
Is this a permutations/combinations question?
well when you do this, everyone gets 2 objects, theres one left, so there will be an uneven amount, what do you think it is

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|dw: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
Consider two cases : 1) exactly two people get 0 objects 2) exactly one person gets 0 object
How many ways are there for Case 1 ?
try again
3 way
Yes
a->0 b->0 c->7 a->7 b->0 c->0 a->0 b->7 c->0 total 3 way
2) 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 = \(2^7-2\) 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 \(3(2^7-2)\)
i still cant understand this, \(2∗2∗2∗2∗2∗2∗2=2^7\)
Remove 3rd person, so now you have only two people : {person1, person2}
You want to distribute the 7 objects to those two people
Look at each object, It can go to either `person1` or `person2` so each object has `2` different states, yes ?
is this equivalent to find the number of whole number soln's of a+b=7
yes it has 2 states
Nope, because the 7 objects in our case are not identical.
ok
Not just the number of objects, but also the type of object also matters
For 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}
looks like you need to add 3 to your solution
ok
is this also one of the way way 1 : Person1 : {o1,06} Person2 : {o2, o3, o4, o5, o7}
|dw:1443029763196:dw|
ran out of room, but the pattern continues
I would write \[3\cdot2^7-3\]
I 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
mathmath 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,
Is 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
Exactly!
oh thnks
which 2 cases did u substract from 2^7
you can't give all the objects to just one person
i 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
Yes, because they are are already accounted in Case1
case 1 is this right a->0 b->0 c->7 a->7 b->0 c->0 a->0 b->7 c->0
Yes
Case1 covers "all objects given to a single person"
ok thnks

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