## 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 ?

1. 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}}

2. abb0t

Is this a permutations/combinations question?

3. 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

4. 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

5. ganeshie8

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

6. ganeshie8

How many ways are there for Case 1 ?

7. ganeshie8

try again

8. mathmath333

3 way

9. ganeshie8

Yes

10. mathmath333

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

11. 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 $$3(2^7-2)$$

12. mathmath333

i still cant understand this, $$2∗2∗2∗2∗2∗2∗2=2^7$$

13. ganeshie8

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

14. ganeshie8

You want to distribute the 7 objects to those two people

15. ganeshie8

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

16. mathmath333

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

17. mathmath333

yes it has 2 states

18. ganeshie8

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

19. mathmath333

ok

20. ganeshie8

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

21. 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}

22. Zarkon

23. mathmath333

ok

24. mathmath333

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

25. anonymous

|dw:1443029763196:dw|

26. anonymous

ran out of room, but the pattern continues

27. Zarkon

I would write $3\cdot2^7-3$

28. 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

29. 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,

30. 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

31. ganeshie8

Exactly!

32. mathmath333

oh thnks

33. mathmath333

which 2 cases did u substract from 2^7

34. ganeshie8

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

35. 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

36. ganeshie8

Yes, because they are are already accounted in Case1

37. mathmath333

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

38. ganeshie8

Yes

39. ganeshie8

Case1 covers "all objects given to a single person"

40. mathmath333

ok thnks