Question 74653: This is a question from my son's homework.
"If you have 4 horses and 4 stalls, how many different combinations can you stall the horses?"
We have tried
1 horse stalled 4 ways. 4 horses x 4 ways is 16. But then I realize that more than one horse can be stalled togeather. So now I have confused myself so much I can't help my son.
Found 2 solutions by venugopalramana, ikleyn:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! This is a question from my son's homework.
"If you have 4 horses and 4 stalls, how many different combinations can you stall the horses?"
We have tried
1 horse stalled 4 ways. 4 horses x 4 ways is 16. But then I realize that more than one horse can be stalled togeather. So now I have confused myself so much I can't help my son.
YOUR ANSWER HOLDS FOR YOUR INITIAL BASIS OF 1 HORSE/1 HOUSE
WELL YOUR DOUBT IS GENUINE.
BUT THAT SHOULD BE A PART OF QUESTION.
WHETHER 1 STALL CAN HOUSE ONLY 1 HORSE OR MORE.
ASSUMING WE CAN PUT 0 TO 4 HORSES IN 1 , THE ANSWER IS AS FOLLOWS
IF WE DENOTE X1,X2,X3,X4 ARE THE NUMBER OF HORSES IN THE 4 HOUSES,
THEN THE PROBLEM IS EQUIVALENT TO FINDING NUMBER OF INTEGRAL
SOLUTIONS FOR THE EQUATION
X1+X2+X3++X4=4 WITH THE CONDITION THAT
X1>=0,X2>=0,X3>=0 AND X4>=0
THIS CAN BE TAKEN AS FINDING THE COEFFICIENT OF X^4 IN THE EXPANSION
(X^0+X^1+X^2+X^3.......)^4 [HERE X^0 SIGNIFIES PUTTING 0 HORSES
IN 1 HOUSE,X^1 MEANS PUTTING 1 HORSE IN 1 HOUSE ETC..POWER OF 4 FOR
THE WHOLE SUM SIGNIFIES 4 HOUSES AS ALL THE 4 HOUSES HAVE SAME
POSSIBILITY.
THOUGH WE DONT NEED TO PUT BEYOND X^4,THERE IS NO HARM IN PUTTING
THOSE TERMS,AS WE ARE FINALLY TAKING COEFFICIENT OF X^4 ONLY,
AND THERE IS AN ADVANTAGE TO BE HAD AS YOU CAN SEE BELOW
WE KNOW THAT
1/(1-X)=1+X+X^2+X^3+.......
SO WE HAVE TO FIND COEFFICIENT X^4 IN 1/(1-X)OR IN (1-X)^(-4)
= USING BINOMIAL FORMULA....
=[(-1)^4][(-4)(-5)(-6)(-7)/4!
=840/24=35
Answer by ikleyn(54062)  (Show Source):
You can put this solution on YOUR website! .
This is a question from my son's homework.
"If you have 4 horses and 4 stalls, how many different combinations can you stall the horses?"
We have tried
1 horse stalled 4 ways. 4 horses x 4 ways is 16. But then I realize that more than one horse can be stalled togeather.
So now I have confused myself so much I can't help my son.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @venugopalramana is incorrect.
I came to bring a correct solution.
In this problem, the placement is "one horse in one stall", or "one horse in each stall",
in accordance with common sense.
So, in this problem, we deal with permutations.
They want you to determine the number of permutations.
The number of permutations of 4 objects is 4! = 1*2*3*4 = 24.
So, the number of different possible placements of 4 horses in 4 stalls is 4! = 24. ANSWER
If you want more reasoning, here it is.
The first horse can be placed in any of 4 stalls, giving 4 options.
After that, the second horse can be placed in any of 3 remaining stalls, giving 3 options.
After that, the third horse can be placed in any of 2 remaining stalls, giving 2 options
After that, there is only one remaining stall to place the fourth horse.
It is how 4! = 4*3*2*1 = 24 appears.
At this point, the problem is solved completely.
There is a trap in the problem formulation: it uses the word "combinations",
although in reality, the problem is about permutations.
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