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Is this a permutations/combinations question?

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

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

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

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 \[3\cdot2^7-3\]

Exactly!

oh thnks

which 2 cases did u substract from 2^7

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

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