Combinatorics Fish in a Lake

Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.

How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types?

How many outcomes are possible when 3 of the 10 fish caught are trout?

How many when at least 2 of the 10 are trout?

Any help is greatly appreciated. I want to find out how to do this without simply using brute force. I came to an answer of 555 for the first question using brute force, but I'm not certain that's correct.

#Combinatorics

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

I am unsure whether my answer is correct or not. Someone please comment. But first let me put the full question as it is from its source.

Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.

(a) How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types

(b) How may outcomes are possible when 3 of the 10 fish caught are trout?

(c) How many when at least 2 of the 10 are trout?

Let me add one more to above (please comment if the solution I wrote for this one is correct or not)

(d) What will be the solution to (c) when 5 different types of fishes are caught?

Solution

(a) This is star and bar theorem 2 problem. Let us assume there are 10 stars (representing fishes) put in a row. Now we separate them into 5 groups (representing fish types) by putting 4 bars among them. (Notice that if we put bar before first star on extreme left, it will denote that the first group is empty. Same for bar put after last star and two bars put together without any star between them) So we have total 14 positions (10 stars and 4 bars) out of which we have to select 4 for bars which can be done in $\binom{10+5-1}{5-1}=\binom{14}{4}=4$ ways.

(b) When 3 out of 10 are trout, there can be maximum 4 different types of fishes in remaining 7. With same above logic, this can be done in $\binom{7+4-1}{4-1}=\binom{10}{3}=120$ ways.

(c) When at least 2 out of 10 are trout, there can be maximum 5 different types of fishes in remaining 8. This can be done in $\binom{8+5-1}{5-1}=\binom{12}{4}=495$ ways

(d) This becomes star and bar theorem 1 problem. When there is exactly 2 trouts, there should be 4 types of fishes in remaining 8. So not we order 8 stars in a row. They will have 7 gaps separating them. We put 3 bars in those gaps to separate them into 4 groups. This can be done in $\binom{8-1}{4-1}=\binom{7}{3}=35$ ways. If there are more than 2 trouts, then the remaining 8 fishes must contain 5 types of fishes. This can be done in $\binom{8-1}{5-1}=\binom{8-1}{5-1}=\binom{7}{4}=35$ ways. By product rule we get $35\times 35=1225$

Mahesh Abnave - 4 years, 2 months ago

Are you familiar with stars and bars?

Calvin Lin Staff - 4 years, 4 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$