Tutors Answer Your Questions about Permutations (FREE)
Question 994009: find the permutation of number of ways in which 5 first year student and 5 second year student can be seated on a bench in a examination hall so that no two students of the same class are next to each other ?
Answer by KMST(3791) (Show Source):
Question 994124: Two dice are rolled simultaneously. What is the probability of getting a sum of at least 4?
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! Here are all 36 rolls (count 'em) of a pair of dice:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
Let's eliminate the three in the upper
left corner that don't have a sum of
at least 4, but only a sum of 2 or 3:
(1,3) (1,4) (1,5) (1,6)
(2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
That leaves 33 rolls (count 'em!) that
have sum at least 4.
That's 33 out of 36 or 33/36, which
reduces to 11/12.
Edwin
Question 994100: If the odds against an event happening are 81/55, what is the probability of the event happening?
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! If the odds against an event happening are 81/55, what is the probability of the event happening?

Note: odds against = P(against)/P(for)

P(for) = 55/(55+81) = 55/136 = 0.404..

Cheers,
Stan H.

Question 994001: If I had 5 different paints to paint my front and back door how many combinations can I make
Answer by Gogonati(839) (Show Source):
Question 993967: A container contains 15 diesel engines. The company chooses 7 engines at random, and will not ship the container if any of the engines chosen are defective. Find the probability that a container will be shipped even though it contains 2 defectives if the sample size is 7.
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! A container contains 15 diesel engines. The company chooses 7 engines at random, and will not ship the container if any of the engines chosen are defective. Find the probability that a container will be shipped even though it contains 2 defectives if the sample size is 7.

# of ways to succeed:: 13C7
# of random sets of 7:: 15C7

Ans: 13C7/15C7 = 0.2667

Cheers,
Stan H.

Question 993968: A state lottery game requires that you pick 6 different members from 1 to 43. What is the probability of picking exactly 3 of the 6 numbers correctly?
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! A state lottery game requires that you pick 6 different members from 1 to 43. What is the probability of picking exactly 3 of the 6 numbers correctly?

Ans: 6C3*37C6

Cheers,
Stan H.
Question 993969: A game consists of asking 17 questions to determine a person, place or thing that the other person is thinking of. The first question, which is always, "Is it an animal, vegetable, or mineral?", has three possible answers. All other questions must be answered with "Yes or "No." How many possible objects can be distinguished in this game, assuming that all 17 questions are asked? Are 17 questions enough?
Answer by solver91311(20879) (Show Source):
Question 993972: How many different 7letter words can be made
A. if the first letter must be A, X or T and no letter may be repeated?
B. if repeats are allowed (but the first letter is A, X or T)?
C. How many of the 7letter words (starting with A, X or T) with no repeats end in S?
Found 2 solutions by solver91311, stanbon: Answer by solver91311(20879) (Show Source): Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! How many different 7letter words can be made
A. if the first letter must be A, X or T and no letter may be repeated?
Ans: 3*25*24*23*..*20

B. if repeats are allowed (but the first letter is A, X or T)?
Ans: 3*26^6

C. How many of the 7letter words (starting with A, X or T) with no repeats end in S?
Ans: 3*25*24*..*21*1

Cheers,
Stan H.

Question 993839: An experiment consists of dealing 5 cards from a standard 52 card deck. What is the probability of being dealt a 5, 6, 7, 8, 9, all in the same suit?
Answer by fractalier(2141) (Show Source):
You can put this solution on YOUR website! The total number of five card hands possible is found by
52! / (47!)(5!) = 2598960
And since there are only four hands that satisfy 5,6,7,8,9 of the same suit, we divide the above by four and get
2598960 / 4 = 649740
Thus the odds are 1 deal out of 649,740.
Question 993740: A card is picked from a standard deck of 52 cards, and then a coin is flipped 3 times. How many possible outcomes are there?
Answer by Alan3354(47455) (Show Source):
You can put this solution on YOUR website! A card is picked from a standard deck of 52 cards, and then a coin is flipped 3 times. How many possible outcomes are there?
============
The card is 1 of 52

3 coin flips > 8 possible outcomes.

8*52 = 416
Question 992416: How many odd numbers of four digits each can be formed with the digits 0,1,3,5,6 and 7 without repetition ?
Answer by solver91311(20879) (Show Source):
You can put this solution on YOUR website!
There are 5 ways to select the first digit (only 5 because you cannot use zero as a first digit of a four digit number). Then, because you have to not consider the digit selected for the first digit, but are now allowed to use zero, there are, for each of the 5 ways to select the first digit, 5 ways to select the second digit, for a total of 25 ways to select the first two digits. Then, for each of those 25 ways to select the first two digits there are 4 ways to select the third digit. Then there are 3 ways to select the last digit. 5 X 5 X 4 X 3.
John
My calculator said it, I believe it, that settles it
Question 989986: how many numbers are there between 1000 to 10000 such that it is divisible by 6? ( give that no number repeat itself in the 4 digit number )
Answer by Edwin McCravy(13211) (Show Source):
Question 991001: a circle is divided into five equal sectors with each sector orange, blue or green. the first sector must be orange. how many different pattern are there if adjacent sector are coloured differently
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! There are only 10. Just list them: No
need to calculate them:
O O O O O
B G B B B G B G B B
G B G O G O O B O G
O O O O O
G B G B G G G G G B
B G B O B O O B O G
Edwin
Question 991935: In how many ways can the letters A, B,C,D,E,and F be arranged for a sixletter security code?
Answer by Edwin McCravy(13211) (Show Source):
Question 992154: There are 4 letter boxes in a locality. In how many way can a man drop 3 letters into the letter boxes ?
Answer by Alan3354(47455) (Show Source):
Question 990924: How many real or imaginary 1o letter words can be made from the following letters F,G,P,N,A,G,G,F,U,J
Answer by farohw(153) (Show Source):
You can put this solution on YOUR website!
F,G,P,N,A,G,G,F,U,J
In this 10letter sequence, note that F is repeated twice and G three times.
10!/(2!3!) = 302,400 real or imaginary 10 letter words can be made.
Question 990529: Find the number of ways in which a committee, its chairman and its president can be
selected from a set of n people. A person can be chairman as well as president. Committee
can have any number of people?
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! Choose the chairman any of n ways.
Choose the president any of n ways.
That's n² ways to choose the chairman and president.
For each of the remaining n2 people, we can make two choices, either
1. to choose him/her to serve on the committee, or
2. to choose him/her not to serve on the committee.
That's 2^{n2}, counting the cases where only the chairman and
president make up a 2member committee.
Answer: n^{2}*2^{n2}, assuming it is possible to choose
nobody else but the chairman and president. Otherwise the answer is
n^{2}*(2^{n2}1), where we must have at least one other person. You'll have
to ask your teacher whether or not 2member committees are to be counted.
Edwin
Question 990812: How many numbersbetween 1000&10000 can be formed with the digits 1,3,5,7,9 each digit used only once in each number?
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! {1,3,5,7,9}
Choose the 1st digit any of the 5 ways.
Choose the 2nd digit any of the 4 remaining ways not chosen for the 1st digit.
Choose the 3rd digit any of the 3 remaining ways not chosen for the 1st 2 digits.
Choose the 4th digit either of the 2 remaining ways not chosen for the 1st 3
digits.
That'a 5P4 = 5*4*3*2 = 120 ways
Edwin
Question 990558: The number of ways in which the number 27720 can be split into two factors which are coprimes is
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! 27720 = 2³*3²*5*7*11
There are 5 prime factors. Since the two factors of 27720 must be coprime, if we
choose a 2 for one factor to contain, we must choose 2³ for that factor to
contain. Similarly if we choose a 3 for one factor we must choose 3² for that
factor to contain. So it depends on how many prime factors, not how many times
the prime factors are contained in 22720.
Case 1:
Number of ways we can choose 1 prime factor for the first and 4 prime factors
for the second.
That's 5C1 = 5 ways
1. first factor = 2³, second factor = 3²*5*7*11
2. first factor = 3², second factor = 2³*5*7*11
3. first factor = 5, second factor = 2³*3²*7*11
4. first factor = 7, second factor = 2³*3²*5*11
5. first factor = 11, second factor = 2³*3²*5*7
Case 2:
Number of ways we can choose 2 prime factors for the first and 3 prime factors
for the second.
That's 5C2 = 10 ways
1. first factor = 2³*3², second factor = 5*7*11
2. first factor = 2³*5, second factor = 3²*7*11
3. first factor = 2³*7, second factor = 3²*5*11
4. first factor = 2³*11, second factor = 3²*5*7
5. first factor = 3²*5, second factor = 2³*7*11
6. first factor = 3²*7, second factor = 2³*5*11
7. first factor = 3²*11, second factor = 2³*5*7
8. first factor = 5*7, second factor = 2³*3²*11
9. first factor = 5*11, second factor = 2³*3²*7
10. first factor = 7*11, second factor = 2³*3²*5
Answer 5C1 + 5C2 = 5 + 10 = 15 ways.
Edwin
Question 990789: how many different product of 3 digits can be formed from the digits 2,5 and 7 without repetition ?
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! How many different products of 3 digits??
Only one product of the three digits 2,5, and 7 is possible, that is:
2x5x7 = 2x7x5 = 5x2x7 = 5x7x2  7x2x5 = 7x5x2 = 70.
Maybe you didn't know that "product" means "what you get when you multiply".
Maybe you didn't want any product at all.
Maybe you just wanted to know how many ways 2, 5, and 7 can be rearranged.
That would be 6 ways: 257, 275, 527, 572, 725, 752.
Or, maybe you wanted this kind of a product:
3 ways to choose the first digit
times
2 ways remaining to chose for the second digit, and
times
1 way remaining to choose the third digit.
Which is
3x2x1 or 3! or 3P3 or "number of permutations of 3 things taken 3 at a time"
or 6.
Edwin
Question 990562: The number of ways in which the number 27720 can be split into the product of two factors that are coprimes is
Answer by ikleyn(988) (Show Source):
You can put this solution on YOUR website! .
27720 = 8*9*5*7*11
is the presentation of the number 27720 as the product of five coprime numbers 8, 9, 5, 7, and 11.
So, we have basically 5 coprime numbers, and the question is: in how many ways we can collect some of them into the first factor? (Then the rest of them will
automatically go into the second factor). The order of coprimes in the first factor does not make a difference. (Same as the order in the second factor does not).
It is the same as to ask: how many subsets is there in the set of 5 object?
The answer is: .
Indeed, the empty subset corresponds to the value 1 of the first factor.
The subsets consisting of 1 elements, give the values of 8, 9, 5, 7, and 11 for the first factor.
The subsets consisting of 2 elements give the factors 8*9, 8*5, 8*7, . . . , 7*11.
The subsets . . . . and so on.
Thus the number of ways in which we can construct the first factor is .
But since we do not make the difference between the first and the second factors, we need to divide this number by 2.
So, the answer is: The number of ways in which the number 27720 can be split into the product of two coprime factors is .
Question 990448: 3=a10\3
Answer by Alan3354(47455) (Show Source):
Question 148291: How many ways can a baseball coach choose the first, second and third batters in the lineup of a team of 15 players.
Answer by alnavy70(1) (Show Source):
You can put this solution on YOUR website! In this particular question, the different ways of possible players that a coach can choose from the 1st, 2nd, and 3rd batter. Therefore, using The Rule of Sum we can solve.
For the 1st player he has 15 different ways to choose from
For the 2nd player he has 14 different ways to choose from
For the 3rd player he has 13 different ways to choose from
Therefore,
For the first experiment or the second or the last is 15 + 14 + 13 = 42 possible ways the coach can choose the 1st, 2nd, and 3rd batters.
Question 990311: What is the solution of log3x + 4125 = 3?
Answer by Alan3354(47455) (Show Source):
Question 989990: How many times can 1,4,5,8 be arranged?
Answer by ikleyn(988) (Show Source):
Question 989699: If C(n,3)=C(n,4),then the value of n?????please help me to solve this problem.
Answer by macston(4006) (Show Source):
You can put this solution on YOUR website! .
C(n,3)=n!/(3!(n3)!)
.
C(n,4)=n!/(4!(n4)!)
.
n!/(3!(n3)!)=n!/(4!(n4)!) Divide each side by n!
1/(3!(n3)!)=1/(4!(n4)!) Cross multiply
3!(n3)!=4!(n4)! Divide each side by 3!
(n3)!=4(n4)! Divide each side by (n4)!*(Note below)
n3=4(1)
n3=4
n=7
ANSWER: n=7
.
*NOTE: (n3)!=(n3)(n4)! so (n3)!/(n4)!=(n3)
.
CHECK:
n!/(3!(n3)!)=n!/(4!(n4)!)=
7!/(3!(73!))=7!/(4!(74)!)=
7!/(3!4!)=7!/(4!3!)
(7*6*5)/(3*2*1)=(7*6*5)/(3*2*1)
210/6=210/6
35=35
Question 989393: can you help me solve this please
A resort has 9 empty rooms and 7 travelers who want rooms. In how many ways can the manager assign a room to each traveler given that no two travelers can share a room?
I tried 9!/7!, I tried just 9! and just 7!, and I tried just doing 9 times 7 but everything comes back incorrect.
Answer by solver91311(20879) (Show Source):
Question 984135: The legislature of Puerto Rico consists of a 27member Senate and a 51member House of Representatives.
How many ways are there to choose a group of six members from the Puerto Rican legislature?
Answer by Shai(25) (Show Source):
You can put this solution on YOUR website! Total number of members in the peuto Rico
Legislative assembly =65
Number of ways to choose a 6 member committee
From these 65 members=65C6
Question 988244: Prizes are to be awarded to four different members of a group of 8 people.Find the number of ways in which the prizes can be awarded if there are 2 first prizes and 2 second prizes?
Answer by Shai(25) (Show Source):
You can put this solution on YOUR website! Given an 8 member group
Prizes are to be rendered to 4
of them:2 first prizes and 2 second
Hence for the first prize:
8C2
For the second prizes:
6C2(why 6 because 2 of the members have
Already been bestowed with a prize)
So the total:(8C2 + 6C2)
43
Question 988923: A bowl contains 9 red balls and 9 blue balls. A woman's selects 4 balls at random from the bowl. How many different selections are possible if at least 3 balls must be blue ?
Answer by Shai(25) (Show Source):
You can put this solution on YOUR website! Given:
Total number of red balls : 9
Total number of blue balls : 9
Also a woman selects 4 balls
And also at least 3 balls off the 4 must
Be blue
So first combination is 3 blue and 1 red
(remember the woman picks up 4 balls)
And the second combination is 4 blue and 0
Red(remember at least 3 blue balls)
So the equation is as given below:
(9C3*9C1 + 9C4*9C0)=(84+126)
Your answer is 210
Question 988851: In how many ways could 13 people be divided into five groups containing, respectively, 3, 1, 4, 3, and 2 people?
Answer by Shai(25) (Show Source):
Question 988588: [ ] + [ ] + [ ] = 30
please solve this using
1,3,5,7,9,11,13,15 you can also rpeat these number
Answer by Fombitz(25151) (Show Source):
Question 988367: Five men and three women applied for 2 scholarships. What is the probability that the winners will be a man and a woman?
how do i solve this?
Answer by rfer(15570) (Show Source):
Question 988226: write down the binomial expansion of (1+x)^4 and use the result to evaluate(5/4)^4, correct your answer to three decimal places
Answer by Fombitz(25151) (Show Source):
Question 988258: at a school event, adults are seated in groups of exactly 6 and children are seated in groups of exactly 20. ifthere are same number of adults as children, what is the minimum number of adults attending.
Answer by vleith(2950) (Show Source):
You can put this solution on YOUR website! Find the least common multiple
to do this, find the prime factors in both 6 and 30
6 = 2*3
20 = 2*2*5
Now find the least common multiple
there are 2 twos in 20, and 1 two in 6. SO use the 2 twoa
With 6 adults in a group, you need 10 adult groups to make 60 people
Question 987901: a lock has a 3number combination in which the first number is seven less than the second number and eleven more than the third. if the sum of the three numbers is 53, what is the combination?
Answer by solver91311(20879) (Show Source):
Question 987722: How many different groups can be selected for playing tennis out of 4 ladies and 3 gentlemen,there being one lady and one gentlemen on each side
Answer by Edwin McCravy(13211) (Show Source):
You can put this solution on YOUR website! Each foursome consists of two ladies and 2 gentlemen.
We select the ladies
4 ladies choose 2 = 4C2
and
3 gentlemen choose 2 = 3C2
Answer: 4C2×3C2 = 6×3 = 18
Edwin
Question 987672: In how many ways can 8 persons be arranged together without 1st and 2nd persons coming together?
Answer by josmiceli(13716) (Show Source):
Question 987291: help please ,
Holly wants to choose 5 different decorative tiles out of 8. If she plans to place the 5 tiles in a row, end to end, in how many different ways can she arrange them, from left to right?
show work please thanks
Answer by macston(4006) (Show Source):
Question 987413: A 3digit number will be formed using the numbers 2, 3, 4, 5, and 7. How many are less than 500 if the digits must be distinct?
Answer by macston(4006) (Show Source):
You can put this solution on YOUR website! .
To be less than 500, you have three choices for the hundreds digit, that leaves 4 choices for the tens digits, and three choices for the units digit.
.
3 x 4 x 3 = 36
.
ANSWER: There are 36 possible three digit numbers less than 500 with no repetition of digits.
Question 987285: In how many ways can a 3 digit number be formed using the numbers 09, if each digit is used only one time?
help please
Thank you!
show all work please :)
Answer by MathTherapy(4047) (Show Source):
You can put this solution on YOUR website!
In how many ways can a 3 digit number be formed using the numbers 09, if each digit is used only one time?
help please
Thank you!
show all work please :)
A 3digit number CANNOT have 0 as its first or hundreds digit
Therefore, first digit can be any of 9 digits (1  9)
Second digit can be any of 9 digits (0, or any digit except the first digit)
Third digit can be any of 8 digits (Any digit except the first or second digit)
Number of ways: 9(9)(8), or
Question 987248: Help please, thank you.
the door code to get into a topsecret laboratory is 6 digits. The first 3 digits of the code are all odd and the last 3 digits are all even. Digits can be used more than once. How many possible codes are there to gain access to this laboratory?
Show all work please, thanks! :)
Answer by macston(4006) (Show Source):
You can put this solution on YOUR website! .
There are 5 odd digits.
There are 5 even digits.
There are five choices for each of six digits:
.
5 x 5 x 5 x 5 x 5 x 5 = 5^6 = 15625
.
ANSWER: There are 15625 possible codes.
Question 987214: A board of trustees is made up of 10 people. The board is choosing a chairperson, a secretary and a publicist. If they have already decided upon a chairperson, in how many ways can they choose a secretary and a publicist?
Thanks, please show the work too . Thanks!! :)
Answer by stanbon(69061) (Show Source):
You can put this solution on YOUR website! A board of trustees is made up of 10 people. The board is choosing a chairperson, a secretary and a publicist. If they have already decided upon a chairperson, in how many ways can they choose a secretary and a publicist?

Ans: 9*8 = 72 ways

Cheers,
Stan H.

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
