Saturday

April 19, 2014

April 19, 2014

Number of results: 217

**data management**

I am not doing the problem for you but perhaps I can give you an idea. Let's see. I have to work with M M M A A L S If all letters were different the answer would be 7!, but they are not. Let me try a simpler example Say A A B B A B A B A B B A B A A B B B A A B A B A Hmm six ...
*Saturday, January 19, 2008 at 10:59pm by Damon*

**Math: unting and permutations**

oops, meant "counting" and permutations
*Saturday, October 25, 2008 at 1:02pm by courtney's dad*

**math permutations**

how many distinguishable permutations are there of the letters in the word effective
*Monday, January 28, 2013 at 10:52pm by mary*

**MATH PLEASEEEEE**

U N U S U A L This would be permutations of 7 objects except that we have three identical Us. so permutations of 7 objects = 7! but we have embedded in there the permutations of three objects 3! so I get 7!/3! = 7*6*5*4
*Sunday, January 18, 2009 at 6:03pm by Damon*

**alg2/trig**

Permutations: 5! Permutations with beginning consonants: 3*4!
*Wednesday, April 1, 2009 at 7:58pm by bobpursley*

**Math**

Find the probability that if the letters of the word "parallel" are randomly arranged that the L's will not be together. In class, I'm studying permutations and combinations. The solutions stated the no. of permutations where 3 L's are together is 6!/2!. Could you please ...
*Thursday, October 11, 2012 at 5:47am by Candice*

**Math**

3-letter permutations: = 2x26x26 = 1352 4-letter permutations: = 2x26x26x26 = 35152 total = 36504 If not repetition allowed 3-calls = 2x25x24 = 4 letters = 2x25x24x23 = total = ..
*Friday, March 29, 2013 at 5:10pm by Reiny*

**Math**

If you count the dice as unique, there are 6^3 = 216 possible outcomes There are 42 where the sum is at most 6. p(sum <= 6) = 42/216 = 7/36 1 1 1 1 1 2 1 1 3 1 1 4 and permutations: 24 1 2 2 1 2 3 and permutations: 12 2 2 2 and permutations: 6 If the dice are ...
*Tuesday, September 11, 2012 at 11:44pm by Steve*

**permutations with identical items.**

There are 7 letters, so there are 7! ways to arrange them. But, the two t's and 3 a's are indistinguishable. Since there are 2! ways to arrange the 2 t's, we divide by 2! to get the number of unique permutations. Similarly for the 3 a's. So, the final count is 7!/(2!3!) = 120 ...
*Thursday, January 23, 2014 at 11:56am by Steve*

**Math (please help me Steve)**

Here's a lower bound solution. You may want to investigate other possibilities. The number 7 can be partitioned into 3 ascending non-negative integers in 8 ways, namely: 007# 016 025 034 115# 124 133# 233# There are 3 permutations of partitions (indicated #) with non-distinct ...
*Monday, July 8, 2013 at 6:58pm by MathMate*

**alg2/trig**

I got this question wrong on my test, and I have to correct it :/ How many permutations are there of all the letters in the word "TEXAS" ? The answer to this part is 120, but I have no idea on how to solve the second part without having to write it all down. *How many of these...
*Wednesday, April 1, 2009 at 7:58pm by Chris*

**Algebra 2**

I searched Google under the key words "permutations vs. combinations" to get these possible sources: Combinations and Permutations www.mathsisfun.com/combinatorics/combinations-permutations.htmlCombinations and Permutations. What's the Difference? In English we use the word "...
*Friday, August 10, 2012 at 12:33am by PsyDAG*

**science**

There are 64 possible codon permutations since there are three nucleotides of 4 four possible values (4^3 = 64). That includes duplicates (codons that code for the same amino acid) and unused codons as well. There are fewer effective codon permutations.
*Wednesday, March 18, 2009 at 12:38am by Sean*

**Math**

say we have 8 bands numbered 1 through 8 chance that band 1 is top = 1/8 then chance that band 2 is second = 1/7 then chance that band 3 is third = 1/6 etc so I claim 8! ways (big number, you compute :) Now how many permutations of 8 taken 3 at a time (permutations not ...
*Friday, July 29, 2011 at 6:32pm by Damon*

**algebra ll**

permutations of 4 of ten (if the order of the 4 on the sheet matters) n!/(n-r)! = 10!/6! =10*9*8*7 = 5040 Now if all you care about is the different groups of 4 and not what order on the sheet, you want the combinations of 4 of ten n!/[r!(n-r)!] = permutations/r! =5040 /4*3*2...
*Sunday, October 26, 2008 at 8:36pm by Damon*

**Math**

There are 6 ways to fill first, then 5 ways to fill second, and 4 ways to hand out third prize. So there are 6*5*4 or 120 ways, as Jordan said. Except Jordan meant to say 120 different permutations and not combinations. Permutations imply positioning, while combinations do not.
*Sunday, August 23, 2009 at 7:33pm by Reiny*

**MATH ELE.**

Check this helpful site. http://www.mathsisfun.com/combinatorics/combinations-permutations.html
*Tuesday, August 11, 2009 at 2:03pm by Ms. Sue*

**Business Algebra**

Permutations and Combinations have many real-world applications such as determining the number of ways of being dealt a particular poker hand from a deck of cards at your weekend poker game; or creating the batting order for the baseball team that you manage. Other areas that ...
*Wednesday, June 10, 2009 at 10:18pm by Redbone*

**Math**

http://www.mathsisfun.com/combinatorics/combinations-permutations.html
*Tuesday, November 10, 2009 at 10:44am by Ms. Sue*

**Finite Math**

Please help with this problem. In a student writing contest with 25 entries, 3 essays are selected for first, second, and third place awards and 5 are selected for honorable mention. How many ways can this be done? Thanks. Ok Gen, for the first place there are 25 choices, for ...
*Tuesday, July 25, 2006 at 1:08pm by Gen*

**college math**

6 ways to put an entry in space 1 times 5 ways to put an entry in space 2 times 4 ways to put an entry in space 3 times 3 4 2 5 1 6 looks like 6! = 720 How many permutations of 6 taken 3 at a time? (They are permutations not combinations, because the order in the group of ...
*Wednesday, December 12, 2007 at 9:24pm by Damon*

**Permutation**

Clearly they must alternate boy-girl-boy-girl etc. So the question becomes: How many ways can that be done? Consider chair #1. If a boy goes there, there are 5 possibilities. Then there are 5 for the next chair (any girl), then any of four boys, etc until you get a number of ...
*Sunday, February 27, 2011 at 5:25am by drwls*

**permutations in math**

suppose 15 countries compete in an olympic event. gold, and silver,and bronze medals are awarded. how many different arrangements of winners are there if no country wins more than one medal in this event? show the work. thanks. 15 countries can get the gold 14 countries can ...
*Monday, April 9, 2007 at 4:57pm by dillon*

**Math**

Permutations.....
*Thursday, May 20, 2010 at 7:33pm by Sara*

**heeeeeeeeeeeeelp math**

You were quite right, Andrew. The answer should have been (4!)^4/(4!)² 4! is the number of ways to assemble a single body part. Since there are 4 body parts to assemble, we have (4!)^4 ways to assemble the 4 people where order counts. Since there are 4! permutations of ...
*Saturday, June 1, 2013 at 5:09am by MathMate*

**math ,correction**

Is this correct? consider determining how many possible phone numbers are in an area code (repeated numbers allowed) Is this a combination, a permutation, or neither? I think it would be a combination wouldn't it? Since there are ten number possibilities in each of the three ...
*Friday, July 13, 2007 at 7:47pm by student*

**Alg2**

When there is no repetition of the letters, the number of permutations (order counts) is nPr, taking r letters at a time n!/(n-r)!. If all n letters are taken (as in the present case) to make the word, nPn reduces to n!/(n-n)!=n!/0!=n!/1 = n! However, if there are repetitions ...
*Wednesday, November 30, 2011 at 11:23pm by MathMate*

**statistics**

permutations: nPr = n(n - 1)(n - 2) ... (n - r + 1) = n! / (n - r)!
*Sunday, August 8, 2010 at 6:11pm by Anonymous*

**Pre-Algebra [Permutations]**

1. correct. 2. 6*5*4*3*2*1=6!=you do it.
*Thursday, June 25, 2009 at 3:05pm by bobpursley*

**Pre-Algebra [Permutations]**

123
*Thursday, June 25, 2009 at 3:05pm by Anonymous*

**math**

You can use permutations here 3.2.1=6
*Thursday, September 22, 2011 at 7:45pm by Ammar*

**algebra 2 (permutations & combinations)**

what is C(52,5) ?
*Wednesday, April 10, 2013 at 8:47pm by Reiny*

**math**

how many distinguishable permutations are in the word MUSKETEERS?
*Sunday, April 20, 2008 at 8:31pm by Emily*

**Pre-Algebra [Permutations]**

Thought so, thanks! -MC
*Thursday, June 25, 2009 at 3:05pm by mysterychicken*

**college**

it has to do with choosing, and combinations/permutations though.
*Tuesday, September 21, 2010 at 8:12pm by finite*

**ALGEBRA**

how many permutations are there of the following word. mathematics
*Wednesday, November 16, 2011 at 10:42pm by MONICA*

**math**

if order matters, it's permutations if not, then it's combinations
*Tuesday, September 18, 2012 at 10:53am by Steve*

**Math**

Use either permutations or combinations
*Sunday, April 7, 2013 at 5:24pm by Shaniqua*

**Math**

it's the number of permutations of 3 distinct items: 3! = 6
*Saturday, December 7, 2013 at 11:39am by Steve*

**math**

what is the number of distinguishable permutations of the word hippopotamus?
*Saturday, October 18, 2008 at 10:16pm by kendal*

**math**

this is a permutations 3 cds / 3 cases = 3! = 3 x 2 x 1 = 6 6 different ways
*Tuesday, January 18, 2011 at 1:58pm by helper*

**math**

use permutations and combinations on these types of problems.
*Friday, February 24, 2012 at 6:57pm by L.Bianchessi*

**algebra**

Permutations of 9 objects taken 5 at a time = 9!/(9-5)! where 4! = 4x3x2x1 and 9! = 9x8x7x6x5x4x3x2x1
*Wednesday, March 10, 2010 at 6:31pm by FredR*

**math**

im pretty sure its 256. These r permutations right
*Thursday, January 13, 2011 at 9:51pm by matt*

**math**

how many permutations are there for 4 numbers (0,2,3 and 5) adding up to 25 in 8 turns?
*Monday, February 7, 2011 at 11:23am by al*

**Combinations and Permutations, Steve and Ms. Sue☺**

i mean 8 not a
*Wednesday, April 10, 2013 at 5:44pm by steven*

**math**

Determine the number of permutations of the letter of the word "education"
*Tuesday, June 11, 2013 at 9:42am by michael*

**math**

Determine the number of permutations of the letters of the word "EFFECTIVE"
*Tuesday, June 11, 2013 at 10:10am by michael*

**Combinations and Permutations, Steve and Ms. Sue☺**

1. C 2. B 3. D 4. B 5. D 6. B 7. B 8. B
*Wednesday, April 10, 2013 at 5:44pm by Slim Shady*

**math**

http://www.google.com/search?client=safari&rls=en&q=Combinations+permutations&ie=UTF-8&oe=UTF-8
*Thursday, May 17, 2012 at 2:57pm by PsyDAG*

**Math**

Each permutation can be decomposed in terms of cyclical permutations. The GCM of the cycle lengths is the number of times you need to apply the permutation to get the same result back. This number is, of course, different for each permutation, so we need to find the LCM of ...
*Tuesday, June 18, 2013 at 10:54am by Count Iblis*

**Math**

You are asking for the number of permutations, which in this case would be n! (n factorial). n! = n(n-1)(n-2)...1 Your n = 7. I hope this helps. Thanks for asking.
*Thursday, September 20, 2007 at 8:16pm by PsyDAG*

**probability & statistics**

How many distinguishable permutations of letters are possible in the word COLORADO? is it 8!/3!?
*Wednesday, June 13, 2012 at 9:57pm by Anonymous*

**permutations with identical items. **

how many four letter arrangements are there for the word Mattawa?
*Thursday, January 23, 2014 at 11:56am by Anon*

**permutation - eh?**

a) are the two sequences somehow related? b) what do they have to do with permutations?
*Sunday, April 7, 2013 at 12:03pm by Steve*

**Combinations and Permutations, Steve and Ms. Sue☺**

Are they right? Please help
*Wednesday, April 10, 2013 at 5:44pm by Jman*

**Business Math II**

so I don't have to do any Permutations right? just a simple Pr(E)=n(E)/n(S)
*Sunday, September 23, 2007 at 6:05pm by Dan*

**Math: Permutations and Combinations**

The formula is correct, noting the parentheses, i.e. C(n,r) = n!/((n-r)!r!) Since 0!=1, the expression 8!/((8-0)!*(0)!) = 1
*Friday, August 28, 2009 at 3:57pm by MathMate*

**Math**

i need the answer to finding the number of three letter permutations of the letters l,s,u,v,r
*Wednesday, April 13, 2011 at 9:19pm by Rory*

**Combinations and Permutations, Steve and Ms. Sue☺**

#8. the answer is 6,840 yourwelcome.
*Wednesday, April 10, 2013 at 5:44pm by Sam*

**Combinations and Permutations, Steve and Ms. Sue☺**

#8. the answer is 6,840 yourwelcome.
*Wednesday, April 10, 2013 at 5:44pm by Sam*

**Combinations and Permutations, Steve and Ms. Sue☺**

the answer for number 2- is b and a b 1140
*Wednesday, April 10, 2013 at 5:44pm by steven*

**probability**

The number of different permutations is 26*25*24*23 = 358,800
*Saturday, March 3, 2012 at 10:08pm by drwls*

**college Mathematics**

I am having problems with permutations what is 5P2 The answer I have is 20 am I wrong?
*Friday, April 26, 2013 at 11:44am by Carol *

**college math**

Explain permutations and combinations and the differences between the two. Use examples to illustrate.
*Friday, December 7, 2007 at 8:48pm by Rose*

**math**

Misty - AEB is the same combination as ABE Combinations pay no attention to order, permutations do.
*Monday, January 21, 2008 at 5:41pm by Damon*

**math**

The number of distinguishable permutations that can be formed from the letters of the word PRINT is 120.ITS THAT TRUE or FALSE
*Tuesday, April 19, 2011 at 5:37am by Teii*

**algebra 2 (permutations & combinations)**

from a standard deck of 52 cards, how many different five-card hands can be drawn?
*Wednesday, April 10, 2013 at 8:47pm by grace*

**Data managment math**

I will assume engine up front and caboose at the rear so the different ways relate to the other cars. Now we do not know if the tankers are all picked up together and the flatcars together etc. or not. In practice that would be likely because they would come from different ...
*Tuesday, January 22, 2008 at 7:12pm by Damon*

**algebra 1**

h(3) = -9.8(9) + 30(3) + 1.5 = 3.3 I don't have the foggiest clue how you came up with -195.8 I tried different permutations of the arithmetic, but no luck.
*Friday, February 15, 2013 at 10:44am by Reiny*

**math**

Since "education" can be written as "acdeinotu" a nine letter word without repetition, the number of permutations is 9! = 362880
*Tuesday, June 11, 2013 at 9:42am by MathMate*

**math**

How many distinguishable permutations of letters are possible in the word? BASEBALL a. 20,160 b. 10,080 c. 5040 d. 40,320
*Sunday, May 4, 2008 at 9:57pm by Dana*

**Algebra**

I am doing the amount of permutations in the word trigonometry. For the expression, I wrote 12!/2!2!2!. I got 968003200. Is this correct? If not, should I use parenthesese? Thanks
*Friday, January 30, 2009 at 8:22pm by Caleb*

**math**

-Find the number of distinguishable permutations of the given letters ``AAABBBCCC''. -If a permutation is chosen at random, what is the probability that it begins with at least 2 A's?
*Thursday, September 27, 2012 at 12:44am by bballer*

**Liberal Arts Math **

Find the indicated value: a. 4C2 b. List all of the permutations of {a,b,c,d} when the elements are taken two at a time.
*Saturday, September 29, 2012 at 10:11pm by Dawn*

**algebra**

i need explanation how they solved this one the answer is provided: finding the number o fdistinguishable permutations. (15!)/(4!3!612!) = 6, 306,300
*Monday, August 11, 2008 at 5:40pm by student*

**MATH ELE.**

What is the difference between combination snd permutations in Elementary math??
*Tuesday, August 11, 2009 at 2:03pm by TwG*

**Algebra 2**

there are a total of 10 letters, consisting of 3 e's, 2 t's, 2 w's, s, a, and r. for this problem no of permutations (arrangements) = 10!/(3!2!2!) = 151200
*Wednesday, April 22, 2009 at 9:58pm by Reiny*

**math**

If you put all 720 permutations of the digits 1 through 6 in numerical order from least to greatest (ex. 123456, 123465 etc.) What would the 409 permutation be? How do I do it?
*Monday, October 8, 2007 at 10:00am by anonymous*

**Math**

Just different heights? Or are orders of stacking important? We do not have a clear indication of what "this stack of boxes" is. Review your information on combinations and permutations.
*Tuesday, December 8, 2009 at 10:53am by PsyDAG*

**Math**

How many distinct permutations of the letters of the word OTTAWA being and end with the letter T? I did 6!/2! :S but i get 360. I tried 6P2=30 I am getting it wrong... help please.
*Thursday, March 3, 2011 at 11:15pm by Jacob*

**College Math**

Oops - I can't do permutations properly: it's 25 times too big. That last calculation should have been 9.25240 x (10^11). Sorry!
*Saturday, August 25, 2012 at 2:08am by David Q/R*

**permutations and combinations**

Four men and four women are to be seated alternatively at a round table. In how many ways can this be done?
*Monday, January 2, 2006 at 6:28am by bb*

**Math: Permutations and Combinations**

Need a verification: Combinations of n items taken r at a time: C9n,r)= n!/(n-r)!*r! C(8,0) 8!/(8-0)!*(0)! = 8!/8!*0! = 0 or Answer is 0 or 1?
*Friday, August 28, 2009 at 3:57pm by Dee*

**Maths**

I tried using circular permutations......but the answer didn't tally.Also..............31 is a wrong answer
*Thursday, April 18, 2013 at 10:38am by Mathsfreak*

**Math 157**

# of ways of selecting winning quintuple = permutations of 5 objects = 5! = 5 x 4 x 3 x 2 x 1 probability of winning = n!/(r!(n-r)!) = 44!/(5!39!)
*Wednesday, March 10, 2010 at 2:03pm by FredR*

**Algebra 2**

how do you do permutations??? distinguishable ex. Sweetwater would you put the total number of the letters on top and then put what is repeated on bottom???
*Wednesday, April 22, 2009 at 9:58pm by Christina*

**Permutations/Combinations**

In how many ways can 5 men and 5 women be seated at a round table if two men must not sit together?
*Sunday, February 27, 2011 at 7:50am by Newbie*

**math**

By the way the permutations of n items taken r at a time is n!/(n-r)! and if it does not matter how the r items are arranged in each sub group, then it is the number of "combinations" n! / [ r! (n-r)! ]
*Saturday, August 25, 2012 at 7:53pm by Damon*

**Statistics**

So you want the sequence HHHTT or permutations of that So we have 5 things to arrange, 3 alike of one kind, and 2 alike of another. Number of ways = 5!/(3!2!) = 10
*Tuesday, June 15, 2010 at 5:06pm by Reiny*

**algabra II**

does anyone have the answers to exam 00703600 inequalities, permutations, and probabilities i just need the part on graphing the inequalities
*Wednesday, September 29, 2010 at 7:38pm by kay*

**Pre Calc still confused**

Are the brackets right? I've tried (x+5/x)+(6/x-4)=(-7/x^2-4x), and (x+5)/x + 6/(x-4)=-7/(x^2-4x) and a couple of other permutations without achieving enlightenment.
*Saturday, October 24, 2009 at 5:00pm by jim*

**math**

Rewrite the word as: "ceeeffitv" we see that it is a 9 letter word in which the letters e and f repeat 3 and 2 times respectively. The number of permutations is therefore 9!/(3!2!1!1!1!1!) =30240
*Tuesday, June 11, 2013 at 10:10am by MathMate*

**college math**

since order does not matter here, we are dealing with combinations rather than permutations. No. of outfits C(3,1)xC(6,1)xC(14,7) = 3x6x3432 = 61776
*Saturday, February 21, 2009 at 10:46pm by Reiny*

**math**

First solve for x. x = (c-b)/a Then make a table of all possible a,b, c permutations and calculate the corresponding x. a b c x _____________ -3.-4. 6..-10/3 -3. 6. -4.. 10/3 -4.-3...6.. -9/4 -4..6..-3.. 9/4 6..-3..-4.. -1/6 6..-4..-3.. 1/6 Half of the possibilties lead to a ...
*Thursday, February 25, 2010 at 6:22pm by drwls*

**math**

permutations of four of 10 n!/(n-r)! 10!/6! 10*9*8*7 5040
*Thursday, August 2, 2012 at 3:23am by Damon*

**Math**

For the three-letter set {R, S, T}, find the two-letter combinations. I realize that this deals with combinations and not permutations. I just do not understand how they want this done.
*Saturday, May 8, 2010 at 12:02am by BWB*

**statistics permutations and combinations**

Selecting a committee: There are 7 women and 5 men in a department. How many ways can a committee of 4 be selected if there must be at least 2 women?
*Tuesday, November 15, 2011 at 4:52pm by Celee *

**Math**

What we have here is a question from Statistics called Permutations. A permutation is an ordered arrangement of r objects chosen from n objects. Solution: 5 shirts x 6 slacks x 7 ties = 210 outfits can be made.
*Monday, January 7, 2008 at 4:57pm by Guido*

**math**

There are 2^3 = 8 possibilities (permutations) if the order of appearance is important. HHH HHT HTH HTT TTT THT TTH TTT There are 4 combinations (0,1,2 or 3 tails) The probability of TTT is 1/8
*Sunday, January 15, 2012 at 11:31am by drwls*