# Combinatorial Prob.

• Jan 12th 2012, 03:32 PM
kangta27
Combinatorial Prob.
2 questions.

1. Joe, an avid and properly licensed sportsman, is in his hunting blind when he locates 20 canada geese, 25 mallard ducks, 40 bald eagles, 10 whopping cranes, and 5 flamingos. Joe randomly selects six birds to target. What is the probability that at least one of each species is targeted?

2. Thirty items are arranged in a 6 by 5 array as shown.

Attachment 23252
Calculate the number of ways to form a set of three distinct items such that no two of the selected items are i the same row or same column.
• Jan 12th 2012, 05:09 PM
Soroban
Re: Combinatorial Prob.
Hello, kangta27!

Quote:

$\displaystyle \text{2. Thirty numbers are arranged in a }6 \times 5\text{ array as shown.}$

. . $\displaystyle \begin{array}{|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 \\ \hline 6 & 7 & 8 & 9 & 10 \\ \hline 11 & 12 & 13 & 14 & 15 \\ \hline 16 & 17 & 18 & 19 & 20 \\ \hline 21 & 22 & 23 & 24 & 25 \\ \hline 26 & 27 & 28 & 29 & 30 \\ \hline \end{array}$

$\displaystyle \text{Calculate the number of ways to form a set of three distinct numbers so that}$
$\displaystyle \text{no two of the selected numbers are in the same row or same column.}$

Choose any of the 30 available numbers.
Cross out that number and any number in its row and in its column.
This leaves a $\displaystyle 5\times4$ array of numbers.

Choose any of the 20 available numbers.
Cross out that number and any number in its row and in its column.
This leaves a $\displaystyle 4\times3$ array of numbers.

Choose any of the 12 available numbers.

It seems that there are: .$\displaystyle 30\cdot20\cdot12 \:=\:7200$ ways.

But the order of the selection of the three numbers is not considered.
Hence, we must divide by $\displaystyle 3! = 6$

Answer: .$\displaystyle \dfrac{7200}{6} \,=\,1200$ ways.

• Jan 13th 2012, 06:47 AM
Ridley
Re: Combinatorial Prob.
First question: $\displaystyle P(\text{"At least one of each kind"})=\frac{\binom {20} 1 \binom {25} 1 \binom {40} 1 \binom {10} 1 \binom 5 1 \binom {95} 1}{\binom {100} 6}$
• Sep 8th 2012, 04:42 PM
Shwems
Re: Combinatorial Prob.
So please how did we know that the order is not considered ?