Tutors Answer Your Questions about Permutations (FREE)
Question 473849: a survey indicates that 30% of students play football, 60% play basketball, and 10% play both. If a student is picked at random, the ODDS that he plays either football or basketball but not both is
a. 9 to 1
b. 4 to 1
c. 7 to 3 d. 3 to 2 or e. fifty-fifty
can you explain the method of solving it, Does that make sence?
Click here to see answer by robertb(5830)   |
Question 473849: a survey indicates that 30% of students play football, 60% play basketball, and 10% play both. If a student is picked at random, the ODDS that he plays either football or basketball but not both is
a. 9 to 1
b. 4 to 1
c. 7 to 3 d. 3 to 2 or e. fifty-fifty
can you explain the method of solving it, Does that make sence?
Click here to see answer by ashleighb84(7) |
Question 475183: Dear math teacher,
I am not sure why I am having a pretty hard time with combinations. I really nailed the Permutations section of the chapter but the combinations are not that nice to me.
Here is the problem I am struggling with:
"A has 3 maps and B has 9 maps. Determine the number of ways in which they can exchange maps if each keeps his initial number of maps. "
Here is what I did: 12C3 + 12C9 = 220 + 220 = 440 but that is not the right answer. Thank you for your time and help.
Yours respectfully,
I.
Click here to see answer by scott8148(6628)  |
Question 475196: How many different three-digit combinations can be made by using the numbers 1,3,5,7, and 9 without repetitions if the "right" combination can open a safe? Does a combination lock really use combinations?
Click here to see answer by robertb(5830) |
Question 475525: Dear math teacher,
Would you please explain why n cannot equal -4 and 5 as a solution to the following equation?
nC(n-2) = 10
Here is how I solved it:
{n(n-1)}/2 = 10
(n)^2 - n - 20 = 0
(n-5)times(n+4) = 0
n-5 = 0 and n+4 = 0
n = 5 and n = -4
Now, I know that we cannot have a negative factorial, which is why I only selected n = 5 for the answer to this equation. However, if I were to think deeper and say n cannot equal -4 because that would make the expression (n-5)times(n+4) = 0. But this expression is supposed to be equal to zero. So, let's say instead of -4, I got 6, then the 6 would also be a solution to the equation and then, we would have n = {5, 6}.
My second question is: Can n = 2? I don't think so because if n = 2 then the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10 would also become zero. Therefore, n cannot ever equal 2. Is that the correct reasoning?
My third question is: Can n = 0? LOGICALLY, IF I PLUG-IN ZERO IN {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10, THE ENTIRE EXPRESSION WILL BECOME ZERO. SO CAN n EVER EQUAL ZERO OR NO (OF COURSE HYPOTHETICALLY SPEAKING)?
My forth question is: Can n = 1? Here again, the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10 would become zero.
For some reason my textbook, when solving another problem says n cannot equal to a value that you are canceling out. In our problem, it would be n cannot equal to 2 since I am canceling out (n-2)! from numerator and denominator of the expression {n(n-1)times(n-2)!}/{(2.1)times(n-2)!} = 10
Therefore, in my case for the expression
nC(n-2) = 10
{n(n-1)}/2 = 10
(n)^2 - n - 20 = 0
(n-5)times(n+4) = 0
n-5 = 0 and n+4 = 0
n = 5 and n = -4, and n cannot equal 2 and cannot equal -4. But what is the reason n cannot equal 2 and -4? Can n ever equal 0 and -1?
Thank you so much for reading this long question and thank you for finding the time to answer it.
Yours sincerely,
I.
Click here to see answer by solver91311(24713)  |
Question 475561: Dear math teacher,
I have solved the following problem in two ways. Please let me know, whether each way is correct. Thank you for your time and effort.
1.)
nC15 = nC11
nCr = nCn-r
nC15 = nC(n-15) = nC11
Here is where I need help. Why do we simply "drop" n and C from nC(n-15) = nC11
and say:
n-15 = 11
n=26
2.)
nC15 = nC11
nC15 = nC11 = nC(n-11)
Here again, why do we simply "disregard" n and C in nC15 = nC11 = nC(n-11)
to get
15 = n-11
n=26
Thank you again for your time. Have a wonderful evening.
Yours respectfully,
I.
Click here to see answer by scott8148(6628) |
Question 475581: Dear math teacher,
Would you help me explain how to convert r! to r?
Here is the problem I am talking about:
Given nPr=3024 and nCr=126, find r.
nCr=nPr/(r!)
126=3024/(r!)
r!=24
Here is where I get stuck because I don't know how to take factorial to an integer. The textbook's answer lists r=4 as an answer but I do not understand how they converted r! to r. Would you please explain it to me?
Thank you very much for your help and time.
Sincerely,
I.
Click here to see answer by jim_thompson5910(35256) |
Question 475581: Dear math teacher,
Would you help me explain how to convert r! to r?
Here is the problem I am talking about:
Given nPr=3024 and nCr=126, find r.
nCr=nPr/(r!)
126=3024/(r!)
r!=24
Here is where I get stuck because I don't know how to take factorial to an integer. The textbook's answer lists r=4 as an answer but I do not understand how they converted r! to r. Would you please explain it to me?
Thank you very much for your help and time.
Sincerely,
I.
Click here to see answer by MathLover1(20849)  |
Question 475749: In how many ways can 12 books be divided between A and B so that one may get 9 books and the other student 3 books?
Here is how I did it:
12C3 + 3C3 = 220+1 = 221 ways
OR
12C3 +9C9 = 220 + 1 = 221 ways.
The book's answer is 2*12C9 = 2*12C3 = 440 ways. But why my answer is not correct? Isn' t it true that once student A takes 9 books there are only three books left for the second student. So, that's why my calculation makes sense.
Could you help me explain why the textbook is right? Thank you very much.
I.
Click here to see answer by sudhanshu_kmr(1152)  |
Question 475875: There are 20 green marbles and 10 black ones. How many ways can I select 6 green marbles and 2 black ones. I came up with 1,744,200; but this number seems too large so I'd like to double check :) Thank you SOOOO much !
~Stormy
Click here to see answer by scott8148(6628) |
Question 475863: There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points?
Here is how I did it:
n=10 = points total
r=2 = points required to make a line
10C2 = 45 lines
n=4 = colinear points
r=2 = points required to make a line
4C2= 6 lines made from colinear points
n=6 = points that are not colinear
r=2 = points required to make a line
6C2 = 16 lines made from points which are not colinear
But I am not sure what to do with this now. If I multiply 4C2 with 6C2 I get 15*6 = 90 lines. It does not make sense to multiply. Let's do 15 plus 6 that gives 21 lines are made from 10 points but that also sounds wrong; it's too small.
Can I approach this problem this way?
case 1
n=4 colinear points
each colinear point can make 6 lines; thus 1 linear point/6 lines, thus 4*6 = 24 lines
case 2
n=6 (other points, not alligned)
each not alligned point can make 9 lines, thus 1 non-alligned point/9 lines, thus 6*9 = 54 lines
case 1 + case 2 gives 54+24 = 78 lines can be formed by joining 4 colinear points and 6 not alligned points. But this is not the right answer either.
Would you explain please how to proceed with this problem? Thank you so much for your help.
Yours,
I.
Click here to see answer by scott8148(6628) |
Question 475863: There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points?
Here is how I did it:
n=10 = points total
r=2 = points required to make a line
10C2 = 45 lines
n=4 = colinear points
r=2 = points required to make a line
4C2= 6 lines made from colinear points
n=6 = points that are not colinear
r=2 = points required to make a line
6C2 = 16 lines made from points which are not colinear
But I am not sure what to do with this now. If I multiply 4C2 with 6C2 I get 15*6 = 90 lines. It does not make sense to multiply. Let's do 15 plus 6 that gives 21 lines are made from 10 points but that also sounds wrong; it's too small.
Can I approach this problem this way?
case 1
n=4 colinear points
each colinear point can make 6 lines; thus 1 linear point/6 lines, thus 4*6 = 24 lines
case 2
n=6 (other points, not alligned)
each not alligned point can make 9 lines, thus 1 non-alligned point/9 lines, thus 6*9 = 54 lines
case 1 + case 2 gives 54+24 = 78 lines can be formed by joining 4 colinear points and 6 not alligned points. But this is not the right answer either.
Would you explain please how to proceed with this problem? Thank you so much for your help.
Yours,
I.
Click here to see answer by richard1234(7193)  |
Question 475903: From 6 chemists and 5 biologists, a committee of 7 is to be chosen so as to include 4 chemists. In how many ways can this be done?
Here is how I did it:
11C4 times 7C3 = 8250 but it is the wrong answer. Where did I went wrong? Please help me figure this one out. Thank you very much.
I.
Click here to see answer by jorel1380(3719)  |
Question 476093: Dear math teacher,
I am having difficulties with the following problem:
(a)Find n; (b) n cannot equal to .... integers.
(n+2)Cn = 45
(n+2)(n+1)n!/(2.1)n! = 45 and after canceling n! we get
(n+2)(n+1) = 90
n^2 + 3n - 88 = 0
(n+11)(n-8) = 0
n = -11 and n = 8
My answer for (a) is n = 8 only since factorial is not defined for negative numbers. For (b), our n cannot equal -11 and zero since n was canceled out above. Please let me know if my reasoning is correct.
Thank you very much.
I.
Click here to see answer by jim_thompson5910(35256) |
Question 476308: Dear math teacher,
I am having difficulties with the following problem:
4 times nC2 = (n+2)C3, (a)solve for n; (b) n cannot equal to or must be different from ...integers.
4*n(n-1)*(n-2)!/(n-2)!(2.1) = (n+2)!/(n+2-3)!*(3.2.1) and after canceling out 2(n-2)! we get
2n(n-1) = (n+2)(n+1)n(n-1)!/(n-1)!(3.2.1) and after canceling out n and (n-1)! we get
2(n-1) = (n+2)(n+1)/6
n^2 - 9n + 14 = 0
(n-7)(n-2) = 0
n=7 and n=2
To answer part (a) of the problem, I say the solution is 7 only since (n-2) was canceled out above. For part (b) n cannot equal to 0, 1, and 2 since n canceled out, (n-1) canceled out, and (n-2) term canceled out.
Please let me know, if I reason through this correctly. Thank you very much.
Click here to see answer by Theo(13342)  |
Question 476326: Dear math teacher,
I am having difficulties solving for n in the following problem:
nC12 = nC8
n!/(n-12)!12! = n!/(n-8)!8!
n(n-1)...(n-11)(n-12)!/(n-12)!12! = n(n-1)...(n-7)(n-8)!/(n-8)!8!
(n-8)(n-9)(n-10)(n-11)/12! = 1/8!
n^4-21n^3+110n^2-17n^3+357n^2-1870n+72n^2-1512n+7920 / 12! = 1/8!
Would you please let me know whether I am on the right track? My solving steps seem convoluted like a kidney tubule! Please let me know where I made an error.
Thank you very much.
Yours,
I.
Click here to see answer by Theo(13342) |
Question 476358: Dear math teacher,
I am having difficulties with the following problem:
How many straight lines are determined by n points, no three of which lie in the same straight line?
Here is how I reasoned through the problem:
n = n
n = total number of points
3 = points that are NOT collinear
(n-3) = points that ARE collinear
"no three of which lie in the same straight line" means no three of n points lie in the same straight line; therefore,
A.) 3 points are NOT collinear and make nC3 = n!/((n-3)!3!)lines = n(n-1)(n-2)(n-3)!/(n-3)!3! = n(n-1)(n-2)/3! = n(n^2-3n+2)/3! = n^3-3n^2+2n/3! lines
B.) (n-3) points - are collinear points and make 1 line only. They are also points remaining from n points after the 3 points are selected in nC3 ways as in A.)
Total number of lines created from n points, no three of which lie in the same straight line = 1 line + n(n-1)(n-2)/3!; however, the textbook's answer is n(n-1)/2 lines.
Would you please correct me in this problem and explain to me what does "no three of which lie in the same straight line" truly mean?
Thank you very much for helping me figure this out.
Yours respectfully,
Ivanka
Click here to see answer by MathLover1(20849) |
Question 476358: Dear math teacher,
I am having difficulties with the following problem:
How many straight lines are determined by n points, no three of which lie in the same straight line?
Here is how I reasoned through the problem:
n = n
n = total number of points
3 = points that are NOT collinear
(n-3) = points that ARE collinear
"no three of which lie in the same straight line" means no three of n points lie in the same straight line; therefore,
A.) 3 points are NOT collinear and make nC3 = n!/((n-3)!3!)lines = n(n-1)(n-2)(n-3)!/(n-3)!3! = n(n-1)(n-2)/3! = n(n^2-3n+2)/3! = n^3-3n^2+2n/3! lines
B.) (n-3) points - are collinear points and make 1 line only. They are also points remaining from n points after the 3 points are selected in nC3 ways as in A.)
Total number of lines created from n points, no three of which lie in the same straight line = 1 line + n(n-1)(n-2)/3!; however, the textbook's answer is n(n-1)/2 lines.
Would you please correct me in this problem and explain to me what does "no three of which lie in the same straight line" truly mean?
Thank you very much for helping me figure this out.
Yours respectfully,
Ivanka
Click here to see answer by richard1234(7193) |
Question 476372: There are 10 points in a plane. No three of these points are in a straight line, except 4 points which are all in the same straight line. How many straight lines can be formed by joining the 10 points?
total no. of points = n= 10
r = 2 (because each line requires 2 points to be created)
10C2 = 45 total lines can be made
Question 1: The total points refer to 4 points that are collinear and 6 points that are just anywhere in space? Is that correct?
4 collinear points will form only 1 line instead of the hypothetical 4C2 = 6 lines
Question 2: Will 4 points form 6 lines only when they are located anywhere in space?
Question 3: Why do I need to subtract 4C2 = 6 lines from 10C2? Is it because the 6 lines are already included in 10C2?
total no. of lines is = 45 - 6 + 1 = 40
I subtracted 6 because the total number of lines created by 10 points includes 6 lines, so 6 is like having doubles.
Question 4: Why 6 lines have to be subtracted from 45 lines?
Question 5: Does 1 line added to 45 lines account for the 1 line formed by 4 collinear points?
Question 6: What does this mean in mathematical language? "No three of these points are in a straight line, except 4 points which are all in the same straight line."
Here is the textbook's answer:
Number of lines formed if no 3 of the 10 points were in a straight line = 10C2 = 45.
Question 7: I can hardly understand what it means; to me it's just means total number of lines.
Number of lines formed by 4 points, no 3 of which are collinear=4C2=6
Question 8: I cannot really understand what "no 3 of which are collinear" really mean when translated from the language of mathematics to the word problem language.
Since the four points are collinear they form 1 line instead of 6 lines.
Required number of lines = 45-6+1 = 40 lines.
Click here to see answer by richard1234(7193) |
Question 476730: use the combinations formula to write an expression for the number of line segments that join pairs of verticies on an n-sided polygon.
use your result from part (a) to write formula for the numbber of diagonals of an n-sided convex polygon.
Click here to see answer by stanbon(75887) |
Question 477431: Dear math teacher,
I am having diffulties with the following problem:
Find the number of a) combinations and b) permutations of four letters each that can be made from the letters of the word TENNESSEE.
Here is what I did so far, but got stuck/confused:
4-letter word
_ _ _ _ T E N N E S S E E
n = 9
n (vowels) = 4 = {E, E, E, E} therefore n(vowels) = 1
n (consonants) = 5 = { T, N, N, S, S} thus n (consonants) = 3
a) # of combinations = ?
b) # of permutations = ?
(no repetition allowed of same letters for permutations)
Vowels Consonants
n 1 3
r don't know don't know how to get it
E T N S
4! times 4! = 4C4 times 4! = 576 permutations but the answer is 163 permutations.
E T N S
E T S N
E S T N
Please help me figure out what to do with this problem. I got confused because the question is asking for both combinations and permutations; I also got thrown off because there are so many repeats of the letters. Please let me see how you solve it.
Thank you very much.
Yours respectfully,
Ivanka
Click here to see answer by stanbon(75887) |
Question 477438: Dear math teacher,
I am having the following difficulties with this problem:
"A box contains 7 red cards, 6 white cards and 4 blue cards. How many selections of three cards can be made so that a) all three are red, b) none are red?"
I solved the problem correctly but then I tried to solve it the second way - the shorter and less time consuming way. I was supposed to get the same result but did not. Please let me know, why I did not get the same answer.
Here is what I did by Method 1 for part a):
R W B
n 7 6 4
r 3
7C3 = 35 selections
Here is what I did by Method 1 for part b):
R W B
n 7 6 4
r n = 6+4 = 10
r = 3
10C3 = 10!/(10-3)! = 120 selections.
Method 2 part b)
n (total R, W, B) = 7+6+4 = 17
r = 3
17C3 - 7C3 = 680-35 = 645 selections and this answer does not match the answer obtained by Method 1 for part b). Would you tell me why? Where did I do a mistep? Here is how reasoned it out: Total Number of Ways - Number of Ways to Pull Red Cards = Number of Ways to Get Other Colors
Why would that be wrong to think like that?
Thank you very much for your help.
Respectfully,
Ivanka
Click here to see answer by stanbon(75887) |
Question 477545: Dear math teacher,
I am having the following difficulties with this problem:
"A box contains 7 red cards, 6 white cards and 4 blue cards. How many selections of three cards can be made so that a) all three are red, b) none are red?"
I solved the problem correctly but then I tried to solve it the second way - the shorter and less time consuming way. I was supposed to get the same result but did not. Please let me know, why I did not get the same answer.
Here is what I did by Method 1 for part a):
R W B
n 7 6 4
r 3
7C3 = 35 selections
Here is what I did by Method 1 for part b):
R W B
n 7 6 4
r n = 6+4 = 10
r = 3
10C3 = 10!/(10-3)! = 120 selections.
Method 2 part b)
n (total R, W, B) = 7+6+4 = 17
r = 3
17C3 - 7C3 = 680-35 = 645 selections and this answer does not match the answer obtained by Method 1 for part b). Would you tell me why? Where did I do a mistep? Here is how reasoned it out: Total Number of Ways - Number of Ways to Pull Red Cards = Number of Ways to Get Other Colors
Why would that be wrong to think like that?
Thank you very much for your help.
Respectfully,
Ivanka
Click here to see answer by scott8148(6628) |
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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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