mathmath333
  • mathmath333
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 ?
Mathematics
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • 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
  • 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
  • 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
  • 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
  • arindameducationusc
Combination question?
mathmath333
  • mathmath333
well the fact is that i didnt understand the quora answer
imqwerty
  • imqwerty
nd what about the gmat solution :)
mathmath333
  • mathmath333
also the gmat :)
arindameducationusc
  • arindameducationusc
@phi
mathmath333
  • mathmath333
isnt he offline
arindameducationusc
  • arindameducationusc
yes, but if he comes online, he will definitely see, Hez good!
arindameducationusc
  • arindameducationusc
I am trying....
imqwerty
  • imqwerty
ok 1st can u find the number of ways in which u can distribute the gifts :)consider all cases
mathmath333
  • mathmath333
7P3 ?
imqwerty
  • imqwerty
did u count the possibility in which they don't even get a single gift
mathmath333
  • mathmath333
how to do that
imqwerty
  • imqwerty
m also thinkin that
arindameducationusc
  • arindameducationusc
:)
imqwerty
  • imqwerty
(:
mathmate
  • 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
  • mathmate
*have not succeeded
mathmate
  • 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
  • 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
  • mathmath333
yes u r allowed to give 0 objects i think.
ParthKohli
  • 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
  • mathmath333
ok
ParthKohli
  • ParthKohli
You know what, there's another way to do this.
mathmath333
  • mathmath333
well i just need an understandable way lol
ParthKohli
  • ParthKohli
OK, I'd done it but the net got disconnected.
ParthKohli
  • 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
  • mathmath333
3*(7*(2^6-6))
ParthKohli
  • 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
  • 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
  • ParthKohli
Heyyyyy never mind all that, you know
ParthKohli
  • ParthKohli
OK, unfortunately that's it.
ParthKohli
  • ParthKohli
Have we double-counted?
ParthKohli
  • ParthKohli
@ganeshie8 our saviour
ParthKohli
  • 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
  • mathmath333
answer given =1218
ParthKohli
  • ParthKohli
Still not correct.
ParthKohli
  • ParthKohli
ooo got it
ganeshie8
  • 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
  • ganeshie8
At step2, above, the expression \(2^6\) comes from the fact that each gift has \(2\) possible states
mathmath333
  • mathmath333
that looks simple now
ParthKohli
  • ParthKohli
Much sleeker...
ParthKohli
  • ParthKohli
Haha, looks like mathmath had posted that expression earlier. :P
ganeshie8
  • ganeshie8
Actually it was I who posted it from mathmath's computer..
ParthKohli
  • ParthKohli
...what
mathmath333
  • mathmath333
He is a hacker
ParthKohli
  • ParthKohli
oh ew i missed the sarcasm there >_<
ganeshie8
  • ganeshie8
lol we were doing teamviewer earlier but my headset won't cooperate suddenly... so..
ParthKohli
  • ParthKohli
Why did you subtract six?
mathmath333
  • mathmath333
to substract the possibltties where a person gets exactly 1 gift
ganeshie8
  • 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
  • ParthKohli
Yeah, I mean you can give exactly two gifts to two people, as long as the last guy gets five.
ParthKohli
  • ParthKohli
what I mean is you can give exactly one gift to two people
ganeshie8
  • ganeshie8
Yep, so the answer is wrong. It should be : \(3*7(2^6-2*6) = 1092\)
ganeshie8
  • 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
  • ParthKohli
What am I even saying ugh
ganeshie8
  • ganeshie8
No, keep going..
ParthKohli
  • ParthKohli
No, there are six invalid states only!
ParthKohli
  • ParthKohli
But there are two people, so you're right.
ParthKohli
  • ParthKohli
But why exactly is (1, 1, 5) being counted as invalid?
ganeshie8
  • ganeshie8
Because the question says so : "exactly one person gets one gift"
ParthKohli
  • ParthKohli
no, it says at least one person gets exactly one gift
ParthKohli
  • ParthKohli
which is why we need cases
ganeshie8
  • ganeshie8
Oops! I read the problem wrong, wait
ParthKohli
  • ParthKohli
Hey, I found this problem's solution. http://prntscr.com/8k0bq7
ParthKohli
  • 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
  • ParthKohli
http://prntscr.com/8k0czp
ganeshie8
  • ganeshie8
that is nice
ParthKohli
  • 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
  • ganeshie8
Hmm, wonder how is that related to \(3*7(2^6-6)\)
ParthKohli
  • ParthKohli
oh that's actually right... I evaluated it wrongly the last time
ParthKohli
  • ParthKohli
thanks wolfram
ParthKohli
  • ParthKohli
i was getting 1638 the last time because I forgot there was no 2! for the very last term -_-
ParthKohli
  • ParthKohli
what were you guys doing on teamviewer anyway

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