Question

7 different objects must be divided among 3 people .
In how many ways can this be done if at least one
of them gets exactly one object ?

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{7 different objects must be divided among 3 people .}\hspace{.33em}\\~\\
& \normalsize \text{In how many ways can this be done if at least one}\hspace{.33em}\\~\\
& \normalsize \text{of them gets exactly one object ?} \hspace{.33em}\\~\\
& a.)\ 2484 \hspace{.33em}\\~\\
& b.)\ 1218 \hspace{.33em}\\~\\
& c.)\ 729 \hspace{.33em}\\~\\
& d.)\ \normalsize \text{none of these}
\end{align}}\)

Answer 2

i know hw to do it but u do know the detail solution for this comes if u google u want me to give the link and if u have question i ll answer :)

Answer 3

http://gmatclub.com/forum/1-7-different-objects-must-be-divided-among-3-people-in-how-7409.html

Answer 4

https://www.quora.com/How-many-ways-can-7-different-objects-be-distributed-among-3-people-such-that-at-least-one-of-them-gets-exactly-one-object

Answer 5

Combination question?

Answer 6

well the fact is that i didnt understand the quora answer

Answer 7

nd what about the gmat solution :)

Answer 8

also the gmat :)

Answer 9

@phi

Answer 10

isnt he offline

Answer 11

yes, but if he comes online, he will definitely see, Hez good!

Answer 12

I am trying....

Answer 13

ok 1st can u find the number of ways in which u can distribute the gifts :)consider all cases

Answer 14

7P3 ?

Answer 15

did u count the possibility in which they don't even get a single gift

Answer 16

how to do that

Answer 17

m also thinkin that

Answer 18

:)

Answer 19

(:

Answer 20

I have the following calculation, but have succeeded in making a combinatorial verification yet.
Ways to distribute 7 different objects to 3 different bins = 3^7
(order is not important in each bin)

ways to choose 2 bins out of 3 = 3
Ways to distribute 7 different objects to 2 different bins = 2^7

ways to choose 1 bin out of 3 = 3
ways to distribute 7 different objects to 1 bin = 1^7

So total number of ways to distribute 7 different objects to 3 different bins, with at least one object in each bin
= 3^7-3*2^7-3*1^7
= 2187-3(128)-3*(1)
=1800

Answer 21

*have not succeeded

Answer 22

Oh, I did not read correct, it says at least one of them gets \(exactly\) one object.
I will do a variation of the above and come back.

Answer 23

Hey, I guess I know this.

But I'm not sure if we're allowed to give zero objects to any of them. It depends on the case.

Answer 24

yes u r allowed to give 0 objects i think.

Answer 25

First, we fix the person who gets one gift. That makes us fulfil the initial condition. There are \(3\) ways to choose the person who gets one gift.

Now there are six gifts left as we've given one gift to one person.\[x+y = 6\]has solutions \((0,6), (1,5),(2,4),(3,3), (4,2),(5,1),(6,0) \).

Answer 26

ok

Answer 27

You know what, there's another way to do this.

Answer 28

well i just need an understandable way lol

Answer 29

OK, I'd done it but the net got disconnected.

Answer 30

First consider the case where two people get one object each. First we have to choose two people, in \(\binom{3}{2}\) ways. Then we have to choose 2 objects out of 7 and permute them among these two guys. That can be done in 7P2 ways.

Answer 31

3*(7*(2^6-6))

Answer 32

Then consider the case where only one guy gets one object. First choose the guy who gets that one object. \(3\) ways. Then the other two guys have (0,6) (2,4) (3,3) (4,2) (6,0)

Answer 33

For the first and the last case, we can count it in this way:
\[2 \times \binom{7}{6}\]For the second and the second-to-last case, here's how:\[2 \times \binom{7}{4}\times \binom{3}{2}\]For the third case,\[2\times \binom{7}{3}\times \binom{4}{3}\]

Answer 34

Heyyyyy never mind all that, you know

Answer 35

OK, unfortunately that's it.

Answer 36

Have we double-counted?

Answer 37

@ganeshie8 our saviour

Answer 38

Case 1: Exactly two persons get exactly one object.\[\binom{3}{2}\cdot \binom{7}{2}\cdot 2 \]Case 2: Exactly one person gets exactly one object. Number of ways to choose that person is 3. Now the other two are selected.\[0 ~\& ~6: 2! \cdot \binom{7}{6} \]\[2 ~ \& ~ 4: 2! \cdot \binom{7}{4}\cdot \binom{3}{2}\]\[3 ~\& ~3:\binom{7}{3}\cdot \binom{4}{3 } \]

Answer 39

answer given =1218

Answer 40

Still not correct.

Answer 41

ooo got it

Answer 42

1) Give "one" person "one" gift and forget him.
This can be done in \(3\times 7\) ways. (3 ways to choose 1 person, 7 ways to choose 1 gift)

2) Then the task is to distribute "6" gifts to "2" people such that nobody gets exactly "1" gift :
\(2^6-6\)


Multiply them for the final answer.

Answer 43

At step2, above, the expression \(2^6\) comes from the fact that each gift has \(2\) possible states

Answer 44

that looks simple now

Answer 45

Much sleeker...

Answer 46

Haha, looks like mathmath had posted that expression earlier. :P

Answer 47

Actually it was I who posted it from mathmath's computer..

Answer 48

...what

Answer 49

He is a hacker

Answer 50

oh ew i missed the sarcasm there >_<

Answer 51

lol we were doing teamviewer earlier but my headset won't cooperate suddenly... so..

Answer 52

Why did you subtract six?

Answer 53

to substract the possibltties where a person gets exactly 1 gift

Answer 54

At step2, we have 6 gifts and two people.
we don't want to give exactly "1" gift to anyone.

Actually, we need subtract \(2*6\), I think.
Need to think more...

Answer 55

Yeah, I mean you can give exactly two gifts to two people, as long as the last guy gets five.

Answer 56

what I mean is you can give exactly one gift to two people

Answer 57

Yep, so the answer is wrong. It should be :
\(3*7(2^6-2*6) = 1092\)

Answer 58

because, at step2, we have 6 gifts and 2 people
each person can get "1" gift in 6 ways, so 2*6 total invalid states

Answer 59

What am I even saying ugh

Answer 60

No, keep going..

Answer 61

No, there are six invalid states only!

Answer 62

But there are two people, so you're right.

Answer 63

But why exactly is (1, 1, 5) being counted as invalid?

Answer 64

Because the question says so : "exactly one person gets one gift"

Answer 65

no, it says at least one person gets exactly one gift

Answer 66

which is why we need cases

Answer 67

Oops! I read the problem wrong, wait

Answer 68

Hey, I found this problem's solution. http://prntscr.com/8k0bq7

Answer 69

The solution is pretty much how I did it. The only difference was that this solution first distributed one gift to the guy who had one gift, then distributed it to another guy. What I did was that I distributed that one gift in the end.

Answer 70

http://prntscr.com/8k0czp

Answer 71

that is nice

Answer 72

\[\binom{3}{2}\cdot \binom{7}{2}\cdot 2\]\[+ ~3 \left(2!\cdot \binom{7}{6} + 2!\cdot \binom{7}{4}\cdot \binom{3}{2} + \binom{7}{3}\cdot \binom{4}{3}\right)\]

Answer 73

Hmm, wonder how is that related to \(3*7(2^6-6)\)

Answer 74

oh that's actually right... I evaluated it wrongly the last time

Answer 75

thanks wolfram

Answer 76

i was getting 1638 the last time because I forgot there was no 2! for the very last term -_-

Answer 77

what were you guys doing on teamviewer anyway