q39

[Saumyojit]

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?

3c1*7c1 = fixing one person for one object

2^6 coz of six objects remaining for two persons

21 * 2^6 this is my answer coming.

where am i wrong

 

[Deleted member 4993]

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?

3c1*7c1 = fixing one person for one object

2^6 coz of six objects remaining for two persons

21 * 2^6 this is my answer coming.

where am i wrong

If some person can get 0 object, then ....................

 

[pka]

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?
3c1*7c1 = fixing one person for one object
2^6 coz of six objects remaining for two persons
21 * 2^6 this is my answer coming.

Once again we ask: why do you think your answer is incorrect?
It seems the one person was selected; that person received one of the seven objects; and then you counted the number of functions from the set of six remaining objects to the set of two remaining people. Is that correct?

 

[Dr.Peterson]

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?

3c1*7c1 = fixing one person for one object
2^6 coz of six objects remaining for two persons
21 * 2^6 this is my answer coming.

where am i wrong

Do you not see how you are overcounting? What if two people both get exactly one object?

If you want more discussion, please show the actual problem including options, and tell us what answer you are told is correct, in case that is wrong (or equivalent to yours).

 

[Saumyojit]

Do you not see how you are overcounting? What if two people both get exactly one object?

(b)1218

Yes i see i am overcounting . In , (3c1*7c1)->assume i am selecting person a to have Object B and then in 2^6 i am selecting person b to have object c only . But in another event it may happen so In , (3c1*7c1)->assume i am selecting person b to have Object c and then in 2^6 i am selecting person a to have object b only .

What if two people both get exactly one object? I have counted that in 2^6 but i am overcounting at the same time

 

[Dr.Peterson]

What if two people both get exactly one object? I have counted that in 2^6 but i am overcounting at the same time

So don't overcount.

I solved it by taking two cases: either exactly one gets exactly one, or exactly two get exactly one. And I got the "correct" answer, so it probably is correct.

It's interesting that this time they give numerical choices, not expressions; that would be a hint that something more complicated than your initial work will be needed, and that this time they recognize there will be different approaches.

 

[Saumyojit]

Ok I tried to do it but i came up with the same problemn of overcounting .
Case 1 : Exactly one person one object -->(3c1 * 7c1) * (6c2 * 2c1) * (2^4)
I think case 1 produces duplicate arrangement like this : Person A chooses object 1 , person b chooses object 4,5 and then in 2^4 person b chooses object 6 and person C chooses 2,3,7 .
And in another event same thing might happen .


Case 2 : exactly two person one Object Each : (3c2 * 7c2) *2! -->126


Please tell me ur approach

 

[Dr.Peterson]

I think case 1 produces duplicate arrangement

It's not the case, but your approach to it, that produces duplicates! But you can remove them, as I'll do below by subtraction.

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?

Case 1, two get one object: Choose 2 people (3C2 = 3); choose which object each gets (7P2 = 42); third gets all the rest. Total: 3*42 = 126

Case 2, one gets one object: Choose 1 person (3C1 = 3); choose the one object (7C1 = 7); assign each of remaining 6 objects to one of the two remaining people, but subtract ways to give exactly one to one of them (2^6 - 2*6 = 64 - 12 = 52). Total: 3*7*52 = 1092

Grand total: 126 + 1092 = 1218

 

[Saumyojit]

Choose 1 person (3C1 = 3); choose the one object (7C1 = 7); assign each of remaining 6 objects to one of the two remaining people

3c1 * 7c1 * 2c1 (2c1=assign each of remaining 6 objects to one of the two remaining people)

but subtract ways to give exactly one to one of them (2^6 - 2*6 = 64 - 12 = 52).

this i don't understand.
2^6 counts one object to 5 objects or zero to six objects between two persons . But what is 2*6 . Please explain with some eg

 

[Dr.Peterson]

(2c1=assign each of remaining 6 objects to one of the two remaining people)

Please explain this!

this i don't understand.
2^6 counts one object to 5 objects or zero to six objects between two persons .

What does that mean? I counted, as I said, ways to "assign each of the remaining 6 objects to one of the two remaining people".

But what is 2*6 . Please explain with some eg

Please think! What might I mean by "subtract ways to give exactly one to one of them", and calling that 2*6?

Answer: Choose one person to get only one (2C1 = 2 ways); then choose which object to give that person (6C1 = 6 ways). (The rest automatically go to the other (1 way).)

 

[Saumyojit]

Ok I understood now.
In, One gets one object case you are removing any (1,1,5) type of possibility from 2^6 ensuring exactly one person gets exactly one Object.

2^6- 12...what a approach.
Thanks

 

[sky2rain]

I would approach this directly: here are 4 formats: 1,1,5; 1,2,4; 3,3,1; 2,2,3 to divide the items. Only 1,2,4 formats doesn’t have any double counting scenarios hence C7,1*C6*2*C4,4*P3,3 ways. (That is changing the sequence of handing out doesn’t result in 2 people getting exact same 1 or combo of items just because they lined up differently) By examining the 1,1,5 format, the permutation of giving out of the first two item to first 2 people is p3,2: AB, AC, BA, BC, CA, CB. To avoid double counting such as B5C7A1346 and C7B5A1346, we should reduce them: c7,1*c6,1*c5,5*(c3*2*c1,1). Do the same for 3,3,1 and 2,2,3 format and sum all numbers.