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mathmath333
 one year ago
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
 one year ago
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 ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} & \normalsize \text{7 different objects must be divided among 3 people .}\hspace{.33em}\\~\\ & \normalsize \text{In how many ways can this be done if at least one}\hspace{.33em}\\~\\ & \normalsize \text{of them gets exactly one object ?} \hspace{.33em}\\~\\ & a.)\ 2484 \hspace{.33em}\\~\\ & b.)\ 1218 \hspace{.33em}\\~\\ & c.)\ 729 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i 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
 one year ago
Best ResponseYou've already chosen the best response.0http://gmatclub.com/forum/17differentobjectsmustbedividedamong3peopleinhow7409.html

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Combination question?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1well the fact is that i didnt understand the quora answer

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0nd what about the gmat solution :)

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0yes, but if he comes online, he will definitely see, Hez good!

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I am trying....

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0ok 1st can u find the number of ways in which u can distribute the gifts :)consider all cases

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0did u count the possibility in which they don't even get a single gift

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2I 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 = 3^73*2^73*1^7 = 21873(128)3*(1) =1800

mathmate
 one year ago
Best ResponseYou've already chosen the best response.2Oh, 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
 one year ago
Best ResponseYou've already chosen the best response.1Hey, 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
 one year ago
Best ResponseYou've already chosen the best response.1yes u r allowed to give 0 objects i think.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1First, 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 \((0,6), (1,5),(2,4),(3,3), (4,2),(5,1),(6,0) \).

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1You know what, there's another way to do this.

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1well i just need an understandable way lol

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK, I'd done it but the net got disconnected.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1First 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.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Then 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
 one year ago
Best ResponseYou've already chosen the best response.1For the first and the last case, we can count it in this way: \[2 \times \binom{7}{6}\]For the second and the secondtolast 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
 one year ago
Best ResponseYou've already chosen the best response.1Heyyyyy never mind all that, you know

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1OK, unfortunately that's it.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Have we doublecounted?

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8 our saviour

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Case 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
 one year ago
Best ResponseYou've already chosen the best response.1answer given =1218

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.21) 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 : \(2^66\) Multiply them for the final answer.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2At step2, above, the expression \(2^6\) comes from the fact that each gift has \(2\) possible states

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1that looks simple now

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Haha, looks like mathmath had posted that expression earlier. :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Actually it was I who posted it from mathmath's computer..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1oh ew i missed the sarcasm there >_<

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2lol we were doing teamviewer earlier but my headset won't cooperate suddenly... so..

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Why did you subtract six?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1to substract the possibltties where a person gets exactly 1 gift

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2At 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
 one year ago
Best ResponseYou've already chosen the best response.1Yeah, I mean you can give exactly two gifts to two people, as long as the last guy gets five.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1what I mean is you can give exactly one gift to two people

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yep, so the answer is wrong. It should be : \(3*7(2^62*6) = 1092\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2because, 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
 one year ago
Best ResponseYou've already chosen the best response.1What am I even saying ugh

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1No, there are six invalid states only!

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1But there are two people, so you're right.

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1But why exactly is (1, 1, 5) being counted as invalid?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Because the question says so : "exactly one person gets one gift"

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1no, it says at least one person gets exactly one gift

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1which is why we need cases

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Oops! I read the problem wrong, wait

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1Hey, I found this problem's solution. http://prntscr.com/8k0bq7

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\binom{3}{2}\cdot \binom{7}{2}\cdot 2\]\[+ ~3 \left(2!\cdot \binom{7}{6} + 2!\cdot \binom{7}{4}\cdot \binom{3}{2} + \binom{7}{3}\cdot \binom{4}{3}\right)\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Hmm, wonder how is that related to \(3*7(2^66)\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1oh that's actually right... I evaluated it wrongly the last time

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1i was getting 1638 the last time because I forgot there was no 2! for the very last term _

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.1what were you guys doing on teamviewer anyway
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