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.
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
|
|
|
