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 ?

- mathmath333

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- mathmath333

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

- anonymous

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 :)

- anonymous

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

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## More answers

- anonymous

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

- arindameducationusc

Combination question?

- mathmath333

well the fact is that i didnt understand the quora answer

- imqwerty

nd what about the gmat solution :)

- mathmath333

also the gmat :)

- arindameducationusc

@phi

- mathmath333

isnt he offline

- arindameducationusc

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

- arindameducationusc

I am trying....

- imqwerty

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

- mathmath333

7P3 ?

- imqwerty

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

- mathmath333

how to do that

- imqwerty

m also thinkin that

- arindameducationusc

:)

- imqwerty

(:

- mathmate

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

- mathmate

*have not succeeded

- mathmate

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.

- ParthKohli

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.

- mathmath333

yes u r allowed to give 0 objects i think.

- ParthKohli

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 \((0,6), (1,5),(2,4),(3,3), (4,2),(5,1),(6,0) \).

- mathmath333

ok

- ParthKohli

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

- mathmath333

well i just need an understandable way lol

- ParthKohli

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

- ParthKohli

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.

- mathmath333

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

- ParthKohli

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)

- ParthKohli

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,\[2\times \binom{7}{3}\times \binom{4}{3}\]

- ParthKohli

Heyyyyy never mind all that, you know

- ParthKohli

OK, unfortunately that's it.

- ParthKohli

Have we double-counted?

- ParthKohli

@ganeshie8 our saviour

- ParthKohli

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.\[0 ~\& ~6: 2! \cdot \binom{7}{6} \]\[2 ~ \& ~ 4: 2! \cdot \binom{7}{4}\cdot \binom{3}{2}\]\[3 ~\& ~3:\binom{7}{3}\cdot \binom{4}{3 } \]

- mathmath333

answer given =1218

- ParthKohli

Still not correct.

- ParthKohli

ooo got it

- ganeshie8

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.

- ganeshie8

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

- mathmath333

that looks simple now

- ParthKohli

Much sleeker...

- ParthKohli

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

- ganeshie8

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

- ParthKohli

...what

- mathmath333

He is a hacker

- ParthKohli

oh ew i missed the sarcasm there >_<

- ganeshie8

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

- ParthKohli

Why did you subtract six?

- mathmath333

to substract the possibltties where a person gets exactly 1 gift

- ganeshie8

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...

- ParthKohli

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

- ParthKohli

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

- ganeshie8

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

- ganeshie8

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

- ParthKohli

What am I even saying ugh

- ganeshie8

No, keep going..

- ParthKohli

No, there are six invalid states only!

- ParthKohli

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

- ParthKohli

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

- ganeshie8

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

- ParthKohli

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

- ParthKohli

which is why we need cases

- ganeshie8

Oops! I read the problem wrong, wait

- ParthKohli

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

- ParthKohli

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.

- ParthKohli

http://prntscr.com/8k0czp

- ganeshie8

that is nice

- ParthKohli

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

- ganeshie8

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

- ParthKohli

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

- ParthKohli

thanks wolfram

- ParthKohli

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

- ParthKohli

what were you guys doing on teamviewer anyway

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