AG2907 wrote:
Not clear, does anyone have another approach?
Sorry to hear that the explanation was not clear. Let me take another stab at it.
I am taking smaller values of the same question for this explanation. The concept can be extended to the question that you’ve posted.
Julian (J), Derek (D) and George (G) together have 4 identical
sugarcane stalks to share. If each of them can be
given any whole number of stalks from O to 4, in how
many different ways can the stalks be distributed?
Note: The question doesn’t mention any constraint, which means J D and G can get 0 stalks or can get all 4.
The sugarcane stalks are represented as X.
Let’s assume that the four stalks are represented as -
X X X X
We will partition (divide & distribute) the stalks among the three individuals using |. To do so we need two partitions as shown below -
X | X X | X
In this representation ⇒ One stalk is distributed to one person, two stalks are distributed to the second person and the last stalk is distributed to the third person.
One of the other possible arrangements could be
X X X | X |
In this representation ⇒ Three stalks are distributed to one person, one stalk is distributed to the second person and zero (no stalk) is distributed to the third person
Similarly we can have other arrangments as shown below
| X | X X X
X | X | X X
..
..
..
..
So essentially the number ways the stalks can be distributed to J, D and G equals various arrangments of the four Xs & two |s (similar finding number of possible arrangements the alphabets in the word MISSISSIPPI)
We have 4 stalks (Xs) & 2 partitions (|s). Hence, the number of ways of arranging the six items is
6! / (2!*4!) = 15.
You can now extend this concept to the problem shared earlier. Hope this helps !
Pls. excuse typos.
AG2907 wrote:
Not clear, does anyone have another approach?
Sorry to hear that the explanation was not clear. Let me take another stab at it.
I am taking smaller values of the same question for this explanation. The concept can be extended to the question that you’ve posted.
Julian (J), Derek (D), and George (G) together have 4 identical
sugarcane stalks to share. If each of them can be
given any whole number of stalks from O to 4, in how
many different ways can the stalks be distributed?
Note: The question doesn’t mention any constraint, which means J D and G can get 0 stalks or can get all 4.
The sugarcane stalks are represented as X.
Let’s assume that the four stalks are represented as -
X X X X
We will partition (divide & distribute) the stalks among the three individuals using |. To do so we need two partitions as shown below -
X | X X | X
In this representation ⇒ One stalk is distributed to one person, two stalks are distributed to the second person, and the last stalk is distributed to the third person.
One of the other possible arrangements could be
X X X | X |
In this representation ⇒ Three stalks are distributed to one person, one stalk is distributed to the second person, and zero (no stalk) is distributed to the third person
Similarly, we can have other arrangments as shown below
| X | X X X
X | X | X X
..
..
..
..
So essentially the number of ways the stalks can be distributed to J, D, and G equals various arrangments of the four Xs & two |s (similar to finding the number of possible arrangements of the alphabet in the word MISSISSIPPI)
We have 4 stalks (Xs) & 2 partitions (|s). Hence, the number of ways of arranging the six items is
6! / (2!*4!) = 15.
You can now extend this concept to the problem shared earlier. Hope this helps!
Pls. excuse typos.
_________________
Want to discuss quant questions and strategies : Join the quant chat group todayPower of Tiny Gains1.01^(365) = 37.780.99^(365) = 0.03