Showing posts with label combinations and permutations. Show all posts
Showing posts with label combinations and permutations. Show all posts
Saturday, May 5, 2012
help desk
2 seats open; 5 candidates running; maybe 1600 people will vote. Two candidates are running as a slate, with posters around town saying "Two seats, two votes."
There will be some bullet-voting. No idea how much.
What's the fewest votes a candidate would need to win if the race is close?
There will be some bullet-voting. No idea how much.
What's the fewest votes a candidate would need to win if the race is close?
Wednesday, July 20, 2011
a company labels its products...
AA2
A company labels its product with a three-character code. Each code consists of two letters (not necessarily different) from the 26 letters of the English alphabet, followed by one digit, as shown above. What is the total number of such codes that are available for labeling the company's product?
I'll post the answer in the comments thread later.
My answer is wrong, and I don't understand why.
Sunday, April 24, 2011
help desk - probability
from Art of Problem Solving Introduction to Counting and Probability by David Patrick, p. 128:
8.2.4 A penny, nickel, and dime are simultaneously flipped. What is the probability that heads are showing on at least 6¢ worth of coins?I can do this by brute force, but I don't see the math.
Friday, April 22, 2011
anonymous and Allison on choosing 2 out of 6 vs choosing 4 out of 6
I mentioned in a comments thread the fact that I'm having a very difficult time grokking the idea that the number of ways you can choose 2 items out of 6 is the same as the number of ways you can choose 4 items out of 6.
Why does 6C2 = 6C4?
Here is Allison's explanation of the specific problem I was asking about.
And here is Anonymous explaining why 6C2 = 6C4:
Why does 6C2 = 6C4?
Here is Allison's explanation of the specific problem I was asking about.
And here is Anonymous explaining why 6C2 = 6C4:
Image you have six balls (red, orange, yellow, green, blue, and purple, for example). You also have a box.
6C2: I have six balls, and I am choosing which two balls to put in the box.
6C4: I have six balls, and I am choosing which four to leave out of the box.
Thus, the combination "I put red and orange in the box," is the same way of splitting up the balls as "I leave yellow, green, blue, and purple out of the box." Counting the ways to put two in the box is the same as counting the ways to leave four out of the box.
Thursday, April 21, 2011
help desk - balls in boxes
from Art of Problem Solving's Introduction to Counting and Probability:
1,2
1,3
1,4
2,3
2,4
3,4
What am I missing?
For the option of (2,1,1), the solution is 6, not 3:
Introduction to Counting & Probability (The Art of Problem Solving)
5.16 How many ways are there to put 4 balls in 3 boxes if:I don't understand the part of the solution that explains how many ways you could put 2 balls in one box, 2 balls in another box, and 0 balls in a third box:
(b) the balls are distinguishable but the boxes are not.
(2,0,0): There are 4C2 = 6 ways to choose the balls for the first box, and the remaining go in the second box. However, the two pairs of balls are interchangeable, so we must divide by 2 to get 6/2 = 3 arrangements.To me, it seems like there would have to be 6 ways to choose 2 balls from balls 1, 2, 3, and 4 for the first box:
1,2
1,3
1,4
2,3
2,4
3,4
What am I missing?
For the option of (2,1,1), the solution is 6, not 3:
There are 4C2 = 6 options for picking the two balls to go in one box, and each of the other two balls goes into its own box.I don't see how (2,0,0) is different from (2,1,1) when the boxes are indistinguishable.
Introduction to Counting & Probability (The Art of Problem Solving)
Monday, April 11, 2011
Sunday, April 10, 2011
help desk - combinations
In the integer 3,589 the digits are all different and increase from left to right. How many integers between 4,000 and 5,000 have digits that are all different and that increase from left to right?
Thursday, April 7, 2011
help desk - how many vertical angles
Here's a problem from The Art of Problem Solving Counting Introduction to Counting and Probability by David Patrick:
How many pairs of vertical angles are formed by five distinct lines that have a common point of intersection?
p. 64
Monday, March 28, 2011
Jo in Oklahoma on exercises vs problems
re: SAT problems, Jo in OKC said...
I asked my daughter today. She took the SAT this fall and got a score in the range you mention.Introduction to Counting & Probability (The Art of Problem Solving)
She said the questions are all routine exercises.
She would agree the AMC questions and AIME questions are problems.
One of her favorite areas of math is counting. :-) I remember covering permutations and combinations in high school math. However, what I learned was just a small fraction of what's covered in Art of Problem Solving's Intro to Counting and Probability course or book.
Wednesday, March 2, 2011
how many different groups of 4?
A school librarian would like to buy subscriptions to 7 new magazines. His budget, however, will allow him to buy only 4 new subscriptions. How many different groups of 4 magazines can he choose?
Counting Principle, Combinations, and Permutations Worksheet (pdf file)
Counting Principle, Combinations, and Permutations Worksheet (pdf file)
Friday, August 6, 2010
help desk - probability
I'm having trouble with some problems in Mary Dolciani's Algebra and Trigonometry: Structure and Method Book 2.
p 749
11. A bag contains 2 red, 4 yellow, and 6 blue marbles. Two marbles are drawn at random. Find the probability of each event.
a. Both are red.
b. Both are yellow.
c. Both are blue
d. One is red and one is yellow.
e. Neither is red.
f. Neither is blue.
For a, b, and c, I can solve this problem using two methods and arrive at the same answer:
a. Both are red.
2/12 x 1/11 = 1/66
OR, using the combination formula, nCr:
sample space (number of ways any 2 marbles can be picked out of 12):
12C2 = 12! ÷ 2!(12-2)!
= 12 x 11 ÷ 2
= 66
ways to pick 2 red out of 2 red:
2C2 = 1
probability of picking 2 red:
1 ÷ 66 or 1/66
Trouble is: when I use both approaches to solve d, I get 2 different answers, and I don't understand why.
d. One is red and one is yellow.
2/12 x 4/11 = 1/6 x 4/11 = 4/66 = 2/33
OR, using combination formula nCr:
12C2 = 66 (number of ways any 2 marbles can be drawn from 12)
2C1 = 2 (number of ways 1 red marble can be drawn from 2 red marbles)
4C1 = 4 (number of ways 1 blue marble can be drawn from 4 blue marbles)
so:
2 x 4 ÷ 66 = 4/33 (probability of drawing 1 red & 1 yellow)
2/33 ≠ 4/33
What am I missing here?
Thank you!
Algebra and Trigonometry: Structure and Method Book 2
p 749
11. A bag contains 2 red, 4 yellow, and 6 blue marbles. Two marbles are drawn at random. Find the probability of each event.
a. Both are red.
b. Both are yellow.
c. Both are blue
d. One is red and one is yellow.
e. Neither is red.
f. Neither is blue.
For a, b, and c, I can solve this problem using two methods and arrive at the same answer:
a. Both are red.
2/12 x 1/11 = 1/66
OR, using the combination formula, nCr:
sample space (number of ways any 2 marbles can be picked out of 12):
12C2 = 12! ÷ 2!(12-2)!
= 12 x 11 ÷ 2
= 66
ways to pick 2 red out of 2 red:
2C2 = 1
probability of picking 2 red:
1 ÷ 66 or 1/66
Trouble is: when I use both approaches to solve d, I get 2 different answers, and I don't understand why.
d. One is red and one is yellow.
2/12 x 4/11 = 1/6 x 4/11 = 4/66 = 2/33
OR, using combination formula nCr:
12C2 = 66 (number of ways any 2 marbles can be drawn from 12)
2C1 = 2 (number of ways 1 red marble can be drawn from 2 red marbles)
4C1 = 4 (number of ways 1 blue marble can be drawn from 4 blue marbles)
so:
2 x 4 ÷ 66 = 4/33 (probability of drawing 1 red & 1 yellow)
2/33 ≠ 4/33
What am I missing here?
Thank you!
Algebra and Trigonometry: Structure and Method Book 2
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