P and C
A Teacher needs to distribute 6 red balls and 6 black balls among 10 students such that each student gets at least one ball .
In how many ways can this be done if all black balls are identical and all red balls are identical ?
Sachin
In how many ways can this be done if all black balls are identical and all red balls are identical ?
Sachin
Labels: SharedPuzzle
posted by Rajesh Lal at
12:29 AM
26 Comments:
I hope all know that P and C stands for permutation and combination
I hope you all know that this is Sachin's homework hahahaha
If each student gets at least one ball then it doesn't matter what the colour is, so the problem comes down to the number of ways 12 balls can be distributed to 10 students so that each gets at least one.
Each of the 12 balls can be distributed to any of 10 students, so there are 10^12 = 1 trillion possible distributions.
Obviously 10 out of this trillion would give all the balls to each student in turn. Someone else can work out how many distributions give all the balls to any two students, then any three, and so on up to any nine.
Subtract all these from 1 trillion and there's your answer.
I know its not exactly puzzle but still a pretty good question for people interested in these type of problems
To the above anonymous,its seems you have misunderstood the question
EACH student should get a ball ie no student is left out
I'll try posting a puzzle next time
and also all the balls are not different so the trillion thing is wrong too
672
wrong
i'll post the answer tomorrow
but i'm sure someone would be able to solve it by then
(its just homework after all)
Maybe 48300?
no but getting closer
(well closer than 672 anyway)
How about 41580?
still wrong
Mh, yeah, I think now I know what I was mistaking... how an idiot...
Anyhow, my last try: 11550
Sorry man
Still wrong
try taking it case by case(there are only two cases)
1) 6 red and 4 black
2) 5 red and 5 black
3) 4 red and 6 black
hjg
12960?
Ragknot
All 12 balls need to distributed
So two cases
a)9 students get 1 ball each and tenth student gets 3 balls
b)8 students get 1 ball each and 2 students get 2 balls each
now the next comment should be the correct answer
1) 6 red and 4 black
2) 5 red and 5 black
3) 4 red and 6 black
4) 6 red and 6 black
5) 6 red and 5 black
6) 5 red and 6 black
12! = 3628800
10! = 3628800
2^12 = 4096 ways
9 ways. It all depends on what kids get 2. Two kids will get two. So it is combinations on kids holding 2 of the same or one of each.
It depends on a lot of stuff
Which student is given 2 or 3 balls
How many of these balls are red or black
In how many way can remaining red and black balls be given to the remaining students
Remember students are different(obviously) while the 6 red balls and 6 black balls are identical
This is gonna be a long explanation
Case 1
9 students get 1 ball each and 1 student gets 3 balls
the student getting 3 balls can be selected in 10 ways and he can get
R 3 2 1 0
B 0 1 2 3
the remaining red and black balls are given to 9 students in 9c3+9c4+9c5+9c6=84+126+126+84=420 ways
So total no of ways for case one=10*420=4200
Case 2
8 students get 1 ball each and 2 student get 2 balls each
the 2 students can be selected in 9c2=45 ways
The 4 balls can be
(R/B)=4/0(1 way),3/1(2 ways),2/2(3 ways),1/3(2 ways),0/4(1 way)
the remaining balls are to be distributed to 8 students
So total no of ways=45*(8c2+2*8c3+3*8c4+2*8c5+8c6)=45*490=22050
final answer=4200+22050=26250
The answer is what the answer is.
999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999
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