Tutors Answer Your Questions about Permutations (FREE)
Question 644453: A manager has a panel of 11 players to choose from . He must pick a team of 6 players to play in a match.
Two of the other players melissa and amy both like the same boy so cannot stand each other . They cannot play together on the same team . How many different teams can be selected if melissa and amy cannot both be on the same team
Click here to see answer by stanbon(75887)  |
Question 644763: Hey guys would be great if you could help me out with the following problem:
4-digit code with 0-9 as possible digits. How many codes are there that
contain the digit "0" exactly twice?
I found that (10^4) - (9^4) = 3439 which is the number of codes with "0" at least once. Any help on how I would go about solving two instances of "0" would be awesome.
Thanks,
Rob
Click here to see answer by josmiceli(19441)  |
Question 645233: A company has six female and eight male employees.How many different 5-person committees can be formed, if the committee must have a group leader and the members at large must consist of two women and two men?
Click here to see answer by stanbon(75887) |
Question 645502: For each of the following relations, determine whether the
relation is reflexive, symmetric, antisymmetric, or transitive.
a) ⊆ Z+ Z+ where a b if a|b (read “a divides b,”
as defined in Section 4.3).
b) is the relation on Z where a b if a|b.
c) For a given universeand a fixed subset C of, define
on () as follows: For A, B ⊆ we have A B if
A ∩ C B ∩ C.
d) On the set A of all lines in R2, define the relation for
two lines 1, 2 by 1 2 if 1 is perpendicular to 2.
e) is the relation on Z where x y if x + y is odd.
f ) is the relation on Z where x y if x − y is even.
g) Let T be the set of all triangles in R2. Define on T by
t1 t2 if t1 and t2 have an angle of the same measure.
h) is the relation on Z Z where (a, b)(c, d) if a ≤ c.
[Note: ⊆ (Z Z) (Z Z).]
6. Which relations in Exercise 5 are partial orders? Which are
equivalence relations?
Click here to see answer by solver91311(24713)  |
Question 645505: For A {a, b, c, d, e}, the Hasse diagram for the poset
(A, ) is shown in Fig. 7.23. (a) Determine the relation matrix
for . (b) Construct the directed graph G (on A) that is
associated with . (c) Topologically sort the poset (A, ).
Click here to see answer by solver91311(24713) |
Question 645715: Question from GRE PowerPrep II Practice Test #2
Quant section 3 of 5
Question 13 or 20
A certain identification code is a list of five symbols: S1,S2,D1,D2,D3. Each of the first 2 symbols must be one of the 26 letters of the English alphabet, and each of the last 3 symbols must be one of the 10 digits. Repeated letters and digits are not allowed.) What is the total number of different identification codes?
Click here to see answer by solver91311(24713) |
Question 645739: Peaceful Travel Agency offers vacation packages. Each vacation package includes a city, a month, and an airline. The agency has 2 cities, 1 month, and 6 airlines to choose from. How many different vacation packages do they offer?
Click here to see answer by stanbon(75887) |
Question 647255: I have a question along the lines of the "how many handshakes" with a twist. I need to find out how many handshakes would occur if 3 people shook hands at once. I think I've figured out a pattern, but I have no idea how to form an equation that would predict the outcome for various numbers. Here's what I've got so far.
I figured out the trick to it, but I can't come up with an equation. I won't try to explain how I did it, but let me know if this makes sense, and what you think.
Ok.. I'll start with 5. If there are 5 people we figured out there are 10 handshakes. I figured out that if there are 5 people, it would be 6+3+1+0+0=10. The takes into account that the first person would be involved with 6 handshakes, the second person would only get credit for 3 (because the rest would count in the first persons), the third person would get credit for 1 (because the rest would count by person 1 and 2), the fourth and fifth persons would not get credit for any because theirs would have been counted by the first 3 people.
Now look at 6 people. There equation would be 10+6+3+1+0+0=20. Same idea. First person gets credit for 10 handshakes, second gets credit for 6...and so on.
7 people equation. 15+10+6+3+1+0+0=35. You get the idea.
Now matter what the last 2 people always come up as zero. If we drop the last two zeros and flip the equation for the above example the equation would be 1+3+6+10+15=35. Now the difference between 1 and 3 is 2, the difference between 3 and 6 is 3, the difference between 6 and 10 is 4. Basically if we use the number of people in the handshake, take away two and then add consective integers to the equation it will always work. Does this make sense?
8 people would look like 1+3+6+10+15+21=56.
Now my question is what should the equation look like?
Thanks for any help you can give me.
Laurie
Click here to see answer by aaronwiz(69) |
Question 647852: Greetings, I would like to know hot to expand the following:(3a-2)^4
At the end of my work I got 3x^4-24x^3-72x^2-96x^1-48
In my lessons and practice homework b always had a variable and in this one it doesn't so I wasn't sure if I a right.
Please confirm and help.
Thank you!
Click here to see answer by Alan3354(69443)  |
Question 648753: john goes to a bookstore with the intent of buying 4 books about American History and then putting them on his new bookshelf. The bookstore he visits has 21 books on the subject . How many different ways are there for John to select 4 books and then arrange then on his bookshelf?
Click here to see answer by lynnlo(4176) |
Question 649391: Combinatorics: How many social security numbers can a state official have under the following schemes? Which of the two schemes will generate enough social security numbers to cover a population of 3,800,000 people?
1. A social security number can begin with any number but two even numbers or two odd numbers cannot be next to each other.
2. All numbers can be used anywhere but without repetitions.
Click here to see answer by Edwin McCravy(20054)  |
Question 650098: Twelve computer monitors are stored in a warehouse. The manager knows that three are defective, but has lost the list. He selects five monitors at random to begin testing.
a. How many different choices of five monitors does he have? 792 is the answer using 12C5 for combinations
b. In how many ways could he happen to select the five monitors that include the defective ones? _________________????
Click here to see answer by ewatrrr(24785)  |
Question 650163: A six by six board has the digits 1, 2, 3, 4, 5 and 6 in each row and in each column. None of these digits appears more than once in each row and in each column.
How many different such combinations are there?
Click here to see answer by lynnlo(4176) |
Question 650931: How many different sets of 4 marbles can you package if you have blue, white, yellow and pink marbles? The answer is 35 but I keep getting 24 using 4!. I have also tried 4C4 and 4P4 using combinations and permutation. but can't figure out how to get 35.
Click here to see answer by ewatrrr(24785) |
Question 651332: twelve people are to travel by three different cars,each of which holds four.Find the number of ways in which the party may be divided if two people refuse to travel in the same car;
(1)Assume the antagonists travel in the same car;
(2)Solve by actually placing the two in different cars.
finally,compare your answers in (1)and (2).
(PLEASE I REALLY NEED THE SOLUTION BEFORE NEXT SATURDAY BECAUSE I'LL BE GOING BACK TO SCHOOL.THANK YOU VERY VERY MUCH.)
Click here to see answer by Edwin McCravy(20054) |
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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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