PG Premier League - Chennai Dream Team

[made_for_iims]

hmmm I think you have forgotten one more thing too...
a != 11 , a+b !=11 , a+b+c !=11

So subtract 1+ 12 + 13C2 cases

Logic is..If he has sold all watches to a....he cant go to b,c,d.....

P.S :- This is along with your and mine assumption that we are considering order....

[exile]

Originally Posted by maverick_srikan

A watch seller sells watches on a street. On any given Day, he visits four customers. He pitches to every customer, but the customer may or maynot buy the watches from him. However, at the end of the day, he manages to sell atleast one watch. In how many ways can he sell his watches to the different customers, if it is known that he has a total of eleven watches with him?

1364 is the answer dude

Anyways, the solution is:

Let the number of watches sold to the First, Second, Third and Fourth guy be a,b,c,d respectively. Let the Watches remaining with the salesman at the end of the day be e.

Hence, a+b+c+d+e=11

The total number of solutions to this is 14c4. However, we have to eliminate one case in which e=11. Hence the answer is 14c4-1=1364

I didnt use the fifth variable.... and hence a+b+c+d =11...

the total no of solutions is 14c3 - 1(all four of them zero..)

what's wrong in this method?

[made_for_iims]

Originally Posted by exile

I didnt use the fifth variable.... and hence a+b+c+d =11...

the total no of solutions is 14c3 - 1(all four of them zero..)

what's wrong in this method?

Actually in this you are assuming that he has sold all the watches... But In this question there is no such mention

[maverick_srikan]

Originally Posted by made_for_iims

hmmm I think you have forgotten one more thing too...
a != 11 , a+b !=11 , a+b+c !=11

So subtract 1+ 12 + 13C2 cases

Logic is..If he has sold all watches to a....he cant go to b,c,d.....

P.S :- This is along with your and mine assumption that we are considering order....

I Dont quite get the logic! Evn If we consider the order,

a=11,b=0,c=0,d=0 and b=11,a=0,c=0 and d=0 are different combinations! We just want the number of ways in which he can sell the watches. How does order come in here?

[made_for_iims]

See like if He is going to customers A,B,C,D in this order.
and he sells all his watches to A... He cant go and approach B,C,D....

so acceptable solutions are D=11 ,C=0,B=0,A=0 but not A=11,B,C,D=0.....
I hope i am able to clear myself

[maverick_srikan]

Originally Posted by made_for_iims

See like if He is going to customers A,B,C,D in this order.
and he sells all his watches to A... He cant go and approach B,C,D....

so acceptable solutions are D=11 ,C=0,B=0,A=0 but not A=11,B,C,D=0.....
I hope i am able to clear myself

Agreed. You say that If he sells all his watches to A, he cannot go to B. But it is not mentioned anywhere that he has to visit all 4 customers. We only want to know the number of ways in which he can sell his watches.

So if he sells all his watches to A, that would count as a legitimate case!

[d2rockstar]

Can anybody post the approach for this question

Q)A group of 24 workers, comprising 8 men,8 women and 8 boys is assigned a piece of work.No two workers in the group have the same capacity to work. Each day, a sub group of workers,comprising 1 man, 1 woman and 1 boy, work together. No sub group with the same composition works for more than a day. In this manner, it takes a total of 512 days for the work to be completed.If all the workers work together for one day, what part of the entire work will be completed?

This question was asked in AIMCAT 0916.

[made_for_iims]

Originally Posted by maverick_srikan

Agreed. You say that If he sells all his watches to A, he cannot go to B. But it is not mentioned anywhere that he has to visit all 4 customers. We only want to know the number of ways in which he can sell his watches.

So if he sells all his watches to A, that would count as a legitimate case!

Its mentioned in the question that he visited 4 customers..

hmmm may be i am trying to solve this more as a puzzle than a mathematics question or a question asked in CAT... Thanks

[exile]

Originally Posted by d2rockstar

Can anybody post the approach for this question

Q)A group of 24 workers, comprising 8 men,8 women and 8 boys is assigned a piece of work.No two workers in the group have the same capacity to work. Each day, a sub group of workers,comprising 1 man, 1 woman and 1 boy, work together. No sub group with the same composition works for more than a day. In this manner, it takes a total of 512 days for the work to be completed.If all the workers work together for one day, what part of the entire work will be completed?

This question was asked in AIMCAT 0916.

on the first day 1w, 1m and 1c works... and so and so forth upto eight days.. on the eigth day all of them would have worked once.... and they would have completed 8/512 or 1/64 part of the work...
if all of them work of them work on the same day... they would complete the same amount of work.. ans: 1/64

hope i'm clear..

p.s: why the hell did i not think of this logic in the exam hall???

[nv_kkd]

@ exile

Is the answer to this question 1/8 or 1/64?