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

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- mathmath333

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

- abb0t

Is this a permutations/combinations question?

- Foxythepirate0984

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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## More answers

- Foxythepirate0984

|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

- ganeshie8

Consider two cases :
1) exactly two people get 0 objects
2) exactly one person gets 0 object

- ganeshie8

How many ways are there for Case 1 ?

- ganeshie8

try again

- mathmath333

3 way

- ganeshie8

Yes

- mathmath333

a->0
b->0
c->7
a->7
b->0
c->0
a->0
b->7
c->0
total 3 way

- ganeshie8

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 \(3(2^7-2)\)

- mathmath333

i still cant understand this,
\(2âˆ—2âˆ—2âˆ—2âˆ—2âˆ—2âˆ—2=2^7\)

- ganeshie8

Remove 3rd person, so now you have only two people : {person1, person2}

- ganeshie8

You want to distribute the 7 objects to those two people

- ganeshie8

Look at each object,
It can go to either `person1` or `person2`
so each object has `2` different states, yes ?

- mathmath333

is this equivalent to find the number of whole number soln's
of a+b=7

- mathmath333

yes it has 2 states

- ganeshie8

Nope, because the 7 objects in our case are not identical.

- mathmath333

ok

- ganeshie8

Not just the number of objects, but also the type of object also matters

- ganeshie8

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}

- Zarkon

looks like you need to add 3 to your solution

- mathmath333

ok

- mathmath333

is this also one of the way
way 1 :
Person1 : {o1,06}
Person2 : {o2, o3, o4, o5, o7}

- anonymous

|dw:1443029763196:dw|

- anonymous

ran out of room, but the pattern continues

- Zarkon

I would write \[3\cdot2^7-3\]

- anonymous

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

- anonymous

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,

- mathmath333

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

- ganeshie8

Exactly!

- mathmath333

oh thnks

- mathmath333

which 2 cases did u substract from 2^7

- ganeshie8

you can't give all the objects to just one person

- mathmath333

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

- ganeshie8

Yes, because they are are already accounted in Case1

- mathmath333

case 1
is this right
a->0
b->0
c->7
a->7
b->0
c->0
a->0
b->7
c->0

- ganeshie8

Yes

- ganeshie8

Case1 covers "all objects given to a single person"

- mathmath333

ok thnks

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