Tutors Answer Your Questions about Permutations (FREE)
Question 1163893: How do you properly use the combination formula,I'm confused on how to use the permutations? The problem I have to solve is "how many ways are there to choose 5 cards from a standard deck of 52 playing cards?"
Click here to see answer by ikleyn(52776)   |
Question 1164276: I dont need answer. I just need i) to understand this problem and need to know how to approach it and ii) where can I find similar problems solved or lessons which will illustrate how to approach this problem on this site. I tried quite a bit on this site under permutations/combination, lessons, 'text-book' on this site. I am unable to find any guidence. Could someone please point me where I can I go on this site
Question: 7*nP3 = 6*n+1P3 |solve for n
If I expand the first term as 7[n!/(n-3)!] = 6[(n+1)!/(n+1)!-3]
am I on right path?
Thak you
Click here to see answer by ikleyn(52776) |
Question 1164356: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!*92!)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by greenestamps(13198)  |
Question 1164356: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!*92!)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by ikleyn(52776) |
Question 1164353: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!92!)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by greenestamps(13198) |
Question 1164353: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!92!)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by ikleyn(52776) |
Question 1164352: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!92!0)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by greenestamps(13198) |
Question 1164352: One of the solved problems on this site under lessons of Combination is:
The Quality Assurance Service of a company has to test a sample of 8 tires selected from among 100 tires. In how many ways 8 tires can be selected for testing from this set of 100 tires?
How far I got:
I punched in the following in my TI calculator:
100!/(8!92!0)
I got overflow error. The solution is given under the problem which I understand very well. I also know the calculators cannot DISPLAY factorials of above 70. Nevertheless, the calculators can handle the calculations far beyound 70! internally and as long as the answer is within the capability of its display, they will show the results...at least from what I believe.
Yet the problem is so easy to calculate by hand and still not calculator.
What gives? I just want to understand. Please help.
Thanks
Click here to see answer by ikleyn(52776) |
Question 1164375: I do understand the relationship between nPr and nCr {( nPr = r!(nCr) } and all. But I am missing something with the following that obviously should be apparent!
How does (nP5) translate into nP4 X (n-4) in the following expression given in the book:
30(nP5/5!) = [30 X nP4 X (n-4)] / 5!
Can somebody show me the intermediate steps...just with the translation of nP5 to as nP4 X (n-4)
Also I am new to this site. How do I insert subscripts in the equations while posting the questions?
Thank you
Click here to see answer by Edwin McCravy(20054)  |
Question 1164376: This is more like math-language problem of mine rather than actual math.
The question starts with "In how many ways n women be seated in a row so that 2 PARTICULAR women will not be next to each other?" Here, my understanding is that those 2 women can be treated as one entity and order does not matter ( Whether they sit Lt and Rt or Rt and Lt), and there is no need to invoke 2! ( I know that 2! is 2 and it does not matter here, but had it been 3 or more then whether it is expressed as 3! or 3 will matter, as you know).
Then it goes on, as part of the solution, "Wih no restrictions, n women may be seated in a row in nPn ways. I f 2 of the n women must always sit next to each other, the number of arrangements = 2!(n-1Pn-1)." Here I think...although I am not sure...2! is appropriate, since there is no "Particular". If there was a "Particular"..2! will not be appropriate and only require 2. (Again please ref above regarding my comment about had it been >2)
And then the solutions goes on to, "Hence the number of ways n women can be seated in a row if 2 PARTICULAR women may never sit together = nPn - 2(n-1Pn-1)." ...2 being the factorial of 2 in question.
What happened it being "Particular" to start with, and how does substracting ( yes... once in a way I can't help using Chauser English for subtraction) arrangements of any two women equates (mathematically!) any two Particular women?
Please dont tell me that I am hung up too much on those two PARTICULAR women !! (=
Just tell me where I am getting turned around. I really need to understand this and get going so I dont insert or not insert n! inappropriately!
Please help
Thanks
Click here to see answer by ikleyn(52776) |
Question 1164460: This is not really a question but a reuquest.
Could someone please upload a LESSON under the broad topic of Combinatorics, under the subheading "Solving for n and r in equations with nCr and nPr."
But if it already exists, please let me know how to navigate to the site.
Thanking you in advance
Click here to see answer by ikleyn(52776) |
Question 1164753: a) In how many different ways can the letters of the word KNOWLEDGE be
arranged in such a way that the vowels always come together?
b) In a group of 6 boys and 4 girls, four children are to be selected.
In how many different ways can they be selected such that at least one boy
should be there?
Click here to see answer by Edwin McCravy(20054) |
Question 1164753: a) In how many different ways can the letters of the word KNOWLEDGE be
arranged in such a way that the vowels always come together?
b) In a group of 6 boys and 4 girls, four children are to be selected.
In how many different ways can they be selected such that at least one boy
should be there?
Click here to see answer by ikleyn(52776) |
Question 1164827: My Question is: Let X = {1,2,...,n}, A = {A⊆X | n ∉ A} and B = {A⊆ X | n ∈ A}. Show that |A|=|B| by (BP).
(I understand what each set means where A is the set where n is not included and B is the set where n is included, I believe I am only counting the subsets of each set and not the elements in each set but im not sure how to do that exactly)
Click here to see answer by solver91311(24713)  |
Question 1165084: Susie has three identical apples, three identical oranges and three identical
pears. She wants to create a straight line arrangement using six of these
pieces of fruit for an art class. In how many different ways can she do this?
Click here to see answer by Seutip(231)  |
Question 1165084: Susie has three identical apples, three identical oranges and three identical
pears. She wants to create a straight line arrangement using six of these
pieces of fruit for an art class. In how many different ways can she do this?
Click here to see answer by ikleyn(52776) |
Question 1165374: A club with 55 college students is doing volunteer work this semester. Each student is volunteering at one of four locations. Here is a summary.
Location Number of students
Nursing home 13
Tutoring center 11
Library 13
Pet shelter 18
Three students from the club are selected at random, one at a time without replacement. What is the probability that none of the three students volunteer at the library?
Do not round your intermediate computations. Round your final answer to three decimal places.
Click here to see answer by ikleyn(52776) |
Question 1166084: Suppose a company needs temporary passwords for the trial of a new time management software. Each password will have one digit, followed by one letter, followed by two digits. The digit
7
will not be used. So, there are
26
letters and
9
digits that will be used. Assume that the digits can be repeated. How many passwords can be created using this format?
Click here to see answer by ikleyn(52776) |
Question 1166200: Dr. Smith is preparing a quiz for his Math class. The quiz contains 6 true/false questions and 7 multiple choice questions with 4 choices each.
In how many ways can the 6 true/false questions be answered?
In how many ways can the 7 multiple choice questions be answered?
In how many ways can all of the questions be answered?
Click here to see answer by ikleyn(52776) |
Question 1166262: 51% of the families in the us had not children under the age of 18; 20% had one child; 19% had two children; 7% had three children; and 3% had four or more children. If a faily is selected at random, find the following probabilities.
a. the family has 2 or 3 children.
Click here to see answer by solver91311(24713) |
Question 1166303: You have 200 different-looking tiles. (Each is a different solid color. You
only have one tile of each color.) You sell trays that are made by lining
up 5 tiles in a row and gluing them to a backing.
A customer orders 4 trays. In how many different ways can you fulfill the
order?
Click here to see answer by greenestamps(13198) |
Question 1166298: You have 200 different-looking tiles. (Each is a different solid color. You
only have one tile of each color.) You sell trays that are made by lining
up 5 tiles in a row and gluing them to a backing. A customer orders 4 trays. In how many different ways can you fulfill the order?
Click here to see answer by greenestamps(13198) |
Question 1167472: A group of eleven seniors, nine juniors, five sophomores, and four freshmen must select a committee of four. How many committees are possible if the committee must contain the following?
(a) one person from each class
(b) any mixture of the classes
(c) exactly two seniors
Click here to see answer by Boreal(15235)  |
Question 1168048: To win at LOTTO in one state, one must correctly select 4 numbers from a collection of 60 numbers (1 through 60). The order in which the selection is made does not matter. How many different selections are possible?
Click here to see answer by ikleyn(52776) |
Question 1168054: Please help me solve this: A group of 6 children are choosing colored pencils to draw a picture. Each child is allowed to select one color. The available colors are green, red, and blue. If the second child refuses to use red pencils and the third child refuses to use blue pencils, then how many ways are there for the children to choose pencils? Assume that there are 12 pencils available of each color, and different children are allowed to choose the same color.
Click here to see answer by greenestamps(13198) |
Question 1168053: A group of 6 children are choosing colored pencils to draw a picture. Each child is allowed to select one color. The available colors are green, red, and blue. If the second child refuses to use red pencils and the third child refuses to use blue pencils, then how many ways are there for the children to choose pencils? Assume that there are 12 pencils available of each color, and different children are allowed to choose the same color.
Click here to see answer by greenestamps(13198) |
Question 1168094: Many casinos have a game called the Big Six Money Wheel, which has 54 slots in which are displayed a Joker, the casino logo, and various dollar amounts, as shown in the table at the top of the next column. Players may bet on the Joker, the casino logo, or one or more dollar denominations. The wheel is spun and if the wheel stops on the same place as the player's bet, the player wins that amount for each dollar bet.
Denomination Number of slots
$40 (Joker) 1
$40 (Casino logo) 1
$20 2
$10 4
$5 7
$2 15
$1 24
Suppose a player bets $17 on the Casino logo denomination. What is the probability of winning this bet?
Click here to see answer by Boreal(15235) |
Question 1168187: You have three different tops, two different hoodies, and four different pairs of jeans hanging in your closet. How many ways can these items be arranged in the closet if the tanktops, hoodies and jeans must be together?
Click here to see answer by VFBundy(438)  |
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035
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