In how many ways can you make this stick?

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vuslouui

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Hello, I am wondering about this question:

Assume that we have 10 marshmallows, 80 bananas, and 50 peaches.
Your task is to first put 3 marshmallows on a stick, then 2 bananas,
and finally 4 peaches. In how many ways can you make this stick?

The stick can be made in 9! / (4! x 3! x 2!) ways. Is this answer correct? The question confuses me a bit.
 
Your problem statement seems to suggest that the items are sequential, as defined. In other words, You CANNOT get banana, marshmallow.

Can we distinguish the peaches from one another?
 
vuslouui said:
Hello, I am wondering about this question:

Assume that we have 10 marshmallows, 80 bananas, and 50 peaches.
Your task is to first put 3 marshmallows on a stick, then 2 bananas,
and finally 4 peaches. In how many ways can you make this stick?

The stick can be made in 9! / (4! x 3! x 2!) ways. Is this answer correct? The question confuses me a bit.
Click to expand...
Looks good to me.
 
vuslouui said:
Assume that we have 10 marshmallows, 80 bananas, and 50 peaches.
Your task is to first put 3 marshmallows on a stick, then 2 bananas,
and finally 4 peaches. In how many ways can you make this stick?
The stick can be made in \(\dfrac{9!}{4! \cdot 3! \cdot 2!}\) ways. Is this answer correct? The question confuses me a bit.
Click to expand...
That answer is correct if we assume that each kind of fruit is indistinguishable from one another in the same kind.
But I wonder why the author went to the trouble to vary the numbers?
 
vuslouui said:
Hello, I am wondering about this question:

Assume that we have 10 marshmallows, 80 bananas, and 50 peaches.
Your task is to first put 3 marshmallows on a stick, then 2 bananas,
and finally 4 peaches. In how many ways can you make this stick?
Click to expand...

I interpreted the problem as you're going to choose 3 marshmallows out of 80 and place them on the stick. Then choose 2 bananas out of 80 and place them on the stick next to the marshmallows. Finally, choose 4 peaches out of 50 and place
them next to the bananas on the stick.

(10 C 3)*(80 C 2)*(50 C 4) =
 
lookagain said:
I interpreted the problem as you're going to choose 3 marshmallows out of 80 and place them on the stick. Then choose 2 bananas out of 80 and place them on the stick next to the marshmallows. Finally, choose 4 peaches out of 50 and place
them next to the bananas on the stick.

(10 C 3)*(80 C 2)*(50 C 4) =
Click to expand...
Really, the problem is just very poorly stated.

Since the number of each item available is given, they are presumed to be distinguishable; since it explicitly says "first ... then ... finally", the order of the types is fixed, but position is relevant. So I'd say (10 P 3)*(80 P 2)*(50 P 4).

On the other hand, nothing is said at all about making the stick itself! Maybe there's only one way to do that.
 
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