1Thanks Subho!
I would be grateful to anyone who could explain me in detail.
1) No. of ways in which 6 diff. toys can be given to 2 children so that all toys donot go to the same child. 62
2) No of ways in which 6 diff. toys may be distributed among 4 children so that each child gets atleast one toy. 7200
3)No. of ways in which 6 diff. toys man be distributed among 4 children (without any restriction on the no. of toys a child may get) 46
My attempts-
A1) The formula that can be used is n!n-rCr-1 as empty groups are not allowed. But I get a wrong answer with this formula.
A2) This question is similar to the above question. We can use the formula n!n-rCr-1 where n=6 and r=4
so answer comes 7200 which is correct (contradictory to the above question).
A3)Here we can use n!n-r+1Cr-1 as empty groups are allowed. Again the answer is wrong.
Please help
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13 Answers
491)let we choose r items from 6 and give it to one kid...automatically the rest 6-r will be given to the other kid...
r items can be given to one from 6 in 6Cr ways...or 6C6-r ways (when we are seeing from other kid's perspective)
so ans is 26C1+26C2+6C3 (think yourself how)!
49we can also do this in this way...
each toy can be divided in 2 ways...
i.e. ultimately we can put one toy either in this group or that group
so for 6 groups,
we have 2x2x2x2x2x2 ways i.e. 64 ways
but these 64 ways have 2 ways in whch either of the 2 kiddos have all the gifts whch is nt desirable..so we subtract 2!!
so ways=64-2 which is in resonance with the previous answer!!
49similarly for 3)
each toy ka 4 ways se sadgati kar sakte hain
i.e. each toy can go to either kid A or B or C or D
so no of ways is 4x4x4x4x4x4=46
1Why cant we do 2) by the method you gave in 1)
i.e. Each toy can be given in 4 ways.....so 6 toys in 46 ways
Out of this there are 4 cases when each boy gets all toys.
so 46 - 4
But the answer is wrong..why??
49i do not quite agree to the second answer given
the answer should anyway be less than that of the third one
i.e. 46 i.e. 212 i.e. 4096
so answer can in no way be 7200
4946-4 is not the correct method..
in this case u have to subtract all the cases in which we have atleast one guy getting 0 toys..
that is NOT 4 but a huge number!!!
49reason: frst we give one toy to kiddo A i.e. in 6 ways....5 toys to go...
next we give one toy to kiddo B i.e. in 5 ways....4 more toys to go...
next we give one toy to kiddo C i.e. in 4 ways....3 more toys to go...
next we give one toy to kiddo D i.e. in 3 ways....2 more left....
now u are at ur discretion to give this 2 toys to four people u'd better not cry u'd better not shout u'd better not pry i'm telling u why..Santa Clause is coming to town!! LOL!!
so that can be done in 42 by logic in quest 3...
so total way is...
6x5x4x3x42
49welcome!
but can u explain now why the second answer is not 46-4??
or put in more precise words can u find the no of ways to distribute the toys such that atleast one of the guys get 0 toys???
30@Swordfish - In your attempt in the second question, n!n-rCr-1 is redundant with values n=6 and r=4. Because, check it becomes something like 2C3!
By the way, are you sure the answer is 7200?
1@Subho---
That way is totally wrong......46 gives 4096.
The answer you got from 6x5x4x3x42 is 5760.
We cannot subtract anything from 4096 as it will lead to a lesser answer than 5760/7200 (which ever is right).
@Ashish...
I'm very sorry the formula is n!n-1Cr-1.
Sorry again.
This formula leads you to the answer 7200
1BTW I figured a new method to do 2)
The possibilities of distribution of toys are-
1 1 1 3 in (6!/3!) x 4! = 2880
1 1 2 2 in (6!/2!2!) x 4! = 4320
adding both you get 7200