Tutors Answer Your Questions about Permutations (FREE)
Question 571543: "what is the slope of a binomial
2x+3"
Answer by Alan3354(21576)   (Show Source):
You can put this solution on YOUR website!"what is the slope of a binomial
2x+3"
-----------
A binomial doesn't have a slope.
Slope is the ratio of changes between 2 variables, such as x and y. The binomial has only x.
Question 571391: A board of directors consisting of 8 males and 4 females must select a 4 member panel, from amung its members, for hiring a CEO. What is the probability that the hiring panel will be all females?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!A board of directors consisting of 8 males and 4 females must select a 4 member panel, from amung its members, for hiring a CEO. What is the probability that the hiring panel will be all females?
---
Number of ways to choose 4 females: 4C4 = 1
--
Number of random groups of 4: 12C4 = 495
---
P(4 females chosen) = 1/495
======
Cheers,
Stan H.
===========
Question 570859: minimum number of individual shoes to be picked up from a dark room(containing 10 pair of shoes). if we have to get at least one proper pair
Found 2 solutions by richard1234, Edwin McCravy:
Answer by richard1234(4789)  (Show Source):
You can put this solution on YOUR website!By Pigeonhole principle, if we have ten boxes corresponding to ten pairs of shoes and we have 11 shoes, two shoes must go into the same box (and hence, make a match).
Answer by Edwin McCravy(6932)  (Show Source):
You can put this solution on YOUR website!
11 individual shoes, for if you picked up only 10, you could possibly pick 10
left shoes and no right shoes, or vice-versa. But if so, the 11th one has to
match one of the other 10.
For your information:
If you pick then the probability of
this many shoes, getting one matched pair is
1 0
2 1/19 = .053
3 3/19 = .158
4 99/323 = .307
5 155/323 = .480
6 211/323 = .653
7 259/323 = .802
8 3815/4199 = .909
9 4071/4199 = .970
10 45933/46189 = .994
11 1
If you pick 6 shoes out of the 20 you are more likely than
not to have picked a matching part.
Edwin
Question 571191: HOW MANY WORDS OF THREE DISTINCT LETTERS CAN BE FORMED FROM THE ALPHABET(A,B,Y,Z)
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!HOW MANY WORDS OF THREE DISTINCT LETTERS CAN BE FORMED FROM THE ALPHABET(A,B,Y,Z)
-----
Ans 4*3*2*1 = 24
=====================
Cheers,
Stan H.
Question 570972: If a jpeg image is 50mb and you are to store 80 copies of it along side with words in a zip,how many words can that zip contain.
Answer by JBarnum(1826)  (Show Source):
You can put this solution on YOUR website!i must be missing something cause words and images are seperate. there is nothing to indicate how many words = how many megabytes, there is not a limit of mb listed. so according to all the lack of any limits and knowlegde of how many bytes for each word the only answer is infinite amount of words per zip.
or not enough information to solve
also the zip doesnt have a limit so it could be 80 imgs w/ 1 terabyte of words each ...lol theres like 3 more definites that need to be addressed before this can be solvable.
zip compression level
zip byte limit
bytes per word
Question 570100: Given the menu below how many different burgers can you make. Meat: single meat, double meat, and triple meat. Cheese: there are 3 different kinds (you can use a max of 3 cheeses). Toppings: there are 13 different kinds of toppings.
Note: a combination is counted by what is on the burger NOT the order of placement.....Wouldn't you solve this by multiplying 3*3*13???
Answer by richard1234(4789) (Show Source):
You can put this solution on YOUR website!It is probably assumed that you can use any of the 13 toppings. Therefore the number of ways is 3*3*(2^13) because for each topping you can either put it on or not put it on, for two possibilities. There are 13 toppings, so 2^13.
Question 570068: how many 7card hands have exactly 3 spades, 3 clubs, and 1 heart can be dealt?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!how many 7card hands have exactly 3 spades, 3 clubs, and 1 heart can be dealt?
-----
Ans: 13C3 * 13C3 * 13C1 = 1063348
=====================
Cheers,
Stan H.
Question 570064: in how many ways can 6 bicycles be parked in a row
Answer by jim_thompson5910(21667) (Show Source):
Question 570067: how many 5- digit numbers can be formed (using 0-9)?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!how many 5- digit numbers can be formed (using 0-9)?
----
1st digit cannot be 0, so 9 ways
Each other digit, 10 ways
---
total number: 9*10^4 = 90000
================================
Cheers,
Stan H.
Question 570070: in how many ways can 7 different card be laid out on a table in a row
Answer by stanbon(48516) (Show Source):
Question 570040: how many ways can 5 paintings be line up on a wall
Answer by jim_thompson5910(21667) (Show Source):
Question 569909: in how many different ways can the symbols *, #, ^ , & , ! be arranged?
Answer by jim_thompson5910(21667) (Show Source):
Question 569688: can you help me with
a + 4 < 10 what is a
Answer by JBarnum(1826) (Show Source):
Question 569544: Find the number of permutations that are possible in letter of the word. i. MATHEMATICS ii. STATISTICS iii. ABRACADABRA
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!Find the number of permutations that are possible in letter of the word.
i. MATHEMATICS::::: 11!/(2!*2!*2!) = 4989600
ii. STATISTICS::::: 10!/(3!*3!2!) = 100800
---
iii. ABRACADABRA
I'll leave that to you.
Cheers,
Stan H.
Question 569378: if there is 3 meats and3 dishes at dinner. how many dfferent combinations of meat and side dishes can be made?
Answer by IWork4Dessert(57)  (Show Source):
You can put this solution on YOUR website!The easiest way to do a problem like this is to draw it out. Let's have x be meat and o be side dishes.
x x x
o o o
Now draw a line from the first x to the first o, then the second o, then the third o. Do the same thing for the other x's. You should get 9 lines. Done!
Question 569184: From the digits 1, 8, and 9, how many two-digit numbers can be formed without repeats?
Answer by Tatiana_Stebko(1060)  (Show Source):
Question 568919: A man claimed not to be the father of a certain child. On the basis of evidence presented, the court felt that this man was twice as likely to be the father as not. The blood type of the child is solely determined by the father, if it did not match the blood type of the child, than he could not be the father. The blood type of the child only occurred in only 10% of the population. The blood test indicated that the man had the same blood type as the child, what is the probability that the man is the father?
Thank You!
Answer by Theo(2967)  (Show Source):
You can put this solution on YOUR website!i interpret this problem as follows:
On the basis of evidence presented, the court felt that this man was twice as likely to be the father as not.
I read this to mean that the probability that he is the father is 2/3 and the probability that he is not the father is 1/3 because 2/3 is twice the value of 1/3.
that's if blood type is not taken into consideration.
if blood type is taken into consideration, then the likelihood that he is the father becomes 10 times greater because only 10% of the population has the same blood type.
since he is already 2 times as likely to be the father than not, then this means that he becomes 10 times as likely as that which mean that he becomes 20 times as likely to be the father as not.
if he is 20 times as likely as not, then i read this to mean that the probability that he is the father is 20 times the probability that he is not the father.
if we let x equal the probability that he is the father, then 1 - x is the probability that he is not the father.
our equation becomes:
x = 20 * (1 - x)
simplify this to get:
x = 20 - 20x
add 20x to both sides of this equation to get:
21x = 20
divide both sides of this equation by 21 to get:
x = 20/21 = .9523809524 which is equal to 95.24% rounded to 2 decimal places.
the probability that he is the father is 2/3 = 66.67% if blood type is not taken into consideration.
the probability that he is the father is 95.24% if blood type is taken into consideration on top of that.
this assumes my assumptions on how to calculate a problem such as this are correct.
not having done one like this before, my assumptions are speculative.
i tried looking at it other ways but came up empty.
Question 192269: Q: There are 12 points in a plane of which 5 are collinear. The number of triangles is?
I think we have to combine C(5,2) (out of 5 collinear, 2 points can form a side of triangle) with the combination of the rest of the points... but i m not getting any answer out of the four options
a)200
b)211
c)210
d)none of these
Answer by sEahors3(4) (Show Source):
Question 568524: a student must answer 3 of the 4 essay questions on a social studies test. how much different selections of questions can be made?
Answer by jim_thompson5910(21667) (Show Source):
Question 568476: Given that P(A) = 0.5, P(B) = 0.6, and P(A and B) = 0.10, determine P(A|B)
Choices are 0.30, 0.17, 0.11, 0.20
Answer by jim_thompson5910(21667) (Show Source):
You can put this solution on YOUR website!P(A|B) = P(A intersect B)/P(B)
P(A|B) = P(A and B)/P(B)
P(A|B) = 0.1/0.6
P(A|B) = 0.1667
which rounds to P(A|B) = 0.17
So the answer is choice B
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If you need more help, email me at jim_thompson5910@hotmail.com
Also, please consider visiting my website: http://www.freewebs.com/jimthompson5910/home.html and making a donation. Thank you
Jim
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Question 568465: A card is selected from a standard deck of 52 playing cards. A standard deck of cards has 12 face cards and four Aces (Aces are not face cards). Find the probability of selecting
· an odd prime number under 10 given the card is a club. (1 is not prime.)
· a Jack, given that the card is not a heart.
· a King given the card is not a face card. ?
Answer by Edwin McCravy(6932) (Show Source):
Question 568478: If P(A or B) = 0.3, P(A) = 0.7 and P(B) = 0.5, determine
P(A and B).
( 0.90
0.45
0.27
0.15
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!If P(A or B) = 0.3, P(A) = 0.7 and P(B) = 0.5, determine
P(A and B).
-----
P(A and B) = P(A) + P(B) - P(A or B)
---
= 0.7 + 0.5 - 0.3
----
= 0.9
================
Cheers,
Stan H.
============
( 0.90
0.45
0.27
0.15
Question 568356: how many bridge hands are possible containing 4 spades,6 diamonds,1 club, and 2 hearts?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!how many bridge hands are possible containing 4 spades,6 diamonds,1 club, and 2 hearts?----
---------------------------
Ans: 13C4*13C6*13C1*13C2 = 1244117160
=============
Cheers,
Stan H.
=============
Question 567752: how many ways can you arrange numbers 1-56, using 6 numbers?
Answer by richard1234(4789) (Show Source):
Question 567840: Oops. I posted this with the wrong email address. Can someone please help.
Which pair has equally likely outcomes? List the letters of the two choices below which have equal probabilities of success, separated by a comma. A standard deck of cards has 52 face cards and four Aces (Aces are not face cards).
A. rolling a sum of 7 on two fair six sided dice
B. drawing a spade out of a standard 52 card deck
C. drawing a queen out of a standard 52 card deck given its a face card.
D. drawing a red card out of a standard 52 card deck
E. rolling a 4, 5 or 6 on one fair six sided die
Answer by Theo(2967) (Show Source):
You can put this solution on YOUR website!drawing a sum of 7 on 2 fair 6 sides dice looks like the probability is equal to 1/6.
there are 6 pairs that can have the sum of 7 out of 36 pairs total. that simplifies to 1/6.
the pairS are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1).
drawing a spade is 13/52 which can be reduced to 1/4.
there are 4 suits in the deck and each of them has 13 cards (12 face plus an ace).
drawing a queen given it's a face card is 4/48 which reduces to 1/12.
4 suits = 52 minus 4 aces = 48.
there are 4 queens in the deck.
drawing a red card is 26/52 which reduces to 1/2.
2 suits are red and 2 suits are black.
rolling a 4, 5, or 6 on 1 fair sided dice is 3/6 which reduces to 1/2.
1 fair die can be 1, 2, 3, 4, 5, or 6.
looks like it's the last 2 that have the same probability.
Question 567607: In a medical Study patients are classified in 8 ways according to whether they have blood type AB+, AB-, A+, A-, B+, B-, O+, O-, and also according to whether their blood pressure is low, normal, or high. Find the number of ways in which a patient can be classified.
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!In a medical Study patients are classified in 8 ways according to whether they have blood type AB+, AB-, A+, A-, B+, B-, O+, O-, and also according to whether their blood pressure is low, normal, or high. Find the number of ways in which a patient can be classified.
----
Ans: 8*3 = 24
======================
Question 567608: If an experiment consists of throwing a die and then drawing a letter at random from the English alphabet, how many points are in the sample space?
-a. How many distinct permutations can be made from the letters of the word columns?
b. How many of these permutations start with the letter m?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!If an experiment consists of throwing a die and then drawing a letter at random from the English alphabet, how many points are in the sample space?
Ans: 6*26
--------------------------
-a. How many distinct permutations can be made from the letters of the word columns?
Ans: 7! = 5040
-------------------
b. How many of these permutations start with the letter m?
Ans: 1*6! = 720
======================
Cheers,
Stan H.
Question 567482: An identification code is to consist of three letters followed by six digits. How many different codes are possible if repetition is permitted?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!An identification code is to consist of three letters followed by six digits. How many different codes are possible if repetition is permitted?
----
Ans: 26^3*10^6 = 17576000000
=================
Cheers,
Stan H.
Question 567481: Hello Tutors. Can I get some help with this homework problem please.
Given that P(A) = 0.6 and P(B) = 0.3, and P(A and B) = 0.18, determine P(B|A
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!Given that P(A) = 0.6 and P(B) = 0.3, and P(A and B) = 0.18, determine P(B|A)
-----------------
P(B | A) = P(B and A)/P(A) = 0.18/0.6 = 0.3
===================
Cheers,
Stan H.
================
Question 566759: If I have three prefixes, 584,585, and 586, how do I calculate the number of telephone numbers that can be made?
Answer by scott8148(5880)  (Show Source):
Question 566618: Suppose that the format for license plates in a certain state is 2 letters followed by 4 numbers.
- How many different plates are there if the letters can be repeated but no two numbers can be the same?
- How many different plates can be made if repetitions of numbers and letters are allowed except that no plate can have four zeros?
Thanks!
Answer by josmiceli(6781)  (Show Source):
You can put this solution on YOUR website!If the letters can be repeated but no two numbers can be the same
if repetitions of numbers and letters are allowed except no plate can have 4 zeros
Question 566299: my question is .. find the coefficient ... x to the 6 power in the expansion of (x^2 -3)^10
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!find the coefficient ... x to the 6 power in the expansion of (x^2 -3)^10
--------------------------------
10C3*(x^2)^3*(-3)^7
------
= 120*x^6*(-2187)
-----
= -262,440
==============
Cheers,
Stan H.
==============
Question 565835: how do you find the surface area of a cylinder?
Answer by ad_alta(170)  (Show Source):
You can put this solution on YOUR website!Imagine cutting the cylinder between the two bases and flattening it out. Then you have a rectangle and two base circles. The rectangle has height 'h' and width the circumference of the circle/base. Let the radius of the base be 'r.' The surface area is 2(pi)r^2+2(pi)rh.
Question 565729: Use counting principle to determine answer to part (a). Assume that each event is equally likely to occur. A couple plans to have children. (a) determine the number of points in the sample space of the possible arrangements of boys and girls.
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!Use counting principle to determine answer to part (a). Assume that each event is equally likely to occur. A couple plans to have children. (a) determine the number of points in the sample space of the possible arrangements of boys and girls.
------
Number(s) are missing from your post.
Cheers,
Stan H.
Question 565727: There are 25 questions on a test with four multiple choice answers each. How many total permutations and combinations of attempts would cover all the possible answers?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!There are 25 questions on a test with four multiple choice answers each. How many total permutations and combinations of attempts would cover all the possible answers?
-------
Each question can be answered in 4 ways.
To # of ways = 4^25
========================================
Cheers,
Stan H.
=========================
Question 565624: how many 4-digit sequences can be formed using the digits 0,1,2,3,4,5,6?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!how many 4-digit sequences can be formed using the digits 0,1,2,3,4,5,6?
Ans: 7P4 = 7!/(3!) = 7*6*5*4 = 840 sequences
==================
Cheers,
Stan H.
==================
Question 565068: How many combinations can you make with 15 apples and 27 banana's?
Answer by Alan3354(21576) (Show Source):
You can put this solution on YOUR website!How many combinations can you make with 15 apples and 27 banana's [sic]?
----------
Why do you have banana's, and not apple's?
-----------
If the apples and bananas (not banana's) are interchangeable, then it's
=1*2 = 2 possiblities
Either apple, banana or
banana, apple
-------
If each specimen is different, then it's
15*27 = 405
Question 564954: The functions f and g are defined by these sets of input and
output values.
g = {(1, 2), ( 2, 4), (5, 5), (6, 2)}
f = {(2, 1), (4, 2), (5, 5), ( 2, 6)}
a. Find g( f (2)).
b. Find f (g(6)).
c. Select any number from the domain of either g or f, and find f (g(x)) or g( f (x)), respectively. Describe what is happening.
Answer by issacodegard(60)  (Show Source):
You can put this solution on YOUR website!The elements of the functions are of the form (input, output). So, find the ordered pair in f that has an input value of 2. It's (2,1). In other words, f(2)=1. So g(f(2))=g(1). Find the element of g that has an input value of 1. It's (1,2), so g(1)=2. Therefore, g(f(2))=2.
The domain of f is {2,4,5,-2} and the domain of g is {1,-2,5,6}, and the ranges of f and g are {2,4,5,-2} and {1,-2,5,6} respectively. All the parts are similar, and knowing the domains and ranges of the functions should help you with understanding the composite functions.
Question 564772: There are 5 blouses, 3 pants, 1 necktie, and 4 pairs of shoes. How many combinations are there?
Answer by ad_alta(170) (Show Source):
Question 564771: How many 6-digit numbers are there in 4,6,2,3,7 and 9 if repetition is not allowed?
Answer by Theo(2967) (Show Source):
You can put this solution on YOUR website!there would be 6! = 6*5*4*3*2*1 = 720
if only 2 numbers (4 and 6), then the number of 2 digit numbers would be equal to 2! = 2 * 1 = 2 and the numbers would be 46, 64.
if only 3 numbers (4 and 6 2), then the number of 3 digit numbers would be equal to 3! = 3 * 2 * 1 = 6 and the numbers would be 462, 426, 642, 624, 246, 264.
720 numbers with 6 digits is much harder to show, but you can see that the formula applies to any number of digits.
Question 564642: Using the letters A B C D E, (letters may not occur more than once) how many three-letter sequences are possible?
How many four-letter and how many five-letter sequences?
Answer by stanbon(48516) (Show Source):
You can put this solution on YOUR website!Using the letters A B C D E, (letters may not occur more than once)
how many three-letter sequences are possible?:::5*4*3 = 60
----------
How many four-letter:::5*4*3*2 = 120
----------
how many five-letter sequences?:::5! = 5*4*3*2*1 = 120
================
Cheers,
Stan H.
=====================
Question 563857: How many diffrent passcodes afe there in a four digit code with number 0 to 9
Answer by fcabanski(385) (Show Source):
You can put this solution on YOUR website!This cannot be answered without additional information.
If repeats are allowed, then it is 10^4 = 10,000.
If repeats aren't allowed then it is 10C4 = 210.
If you need help understanding math so you can solve these problems yourself, then one on one online tutoring is the answer ($30/hr). If you need faster solutions with guaranteed detailed answers, then go with personal problem solving ($3.50-$5.50 per problem). Contact me at fcabanski@hotmail.com
Question 563661: if i have to pick two numbers from 1 to 10, provided the first number can't be greater than the second number. How many possible way I can represent the two number. Please suggest the formula to use?
E.g.
12 - valid
21 - not valid
Answer by josmiceli(6781) (Show Source):
You can put this solution on YOUR website!If the 1st number is 1, the 2nd can be 2-10
If the 1st number is 2, the 2nd can be 3-10
If the 1st number is 3, the 2nd can be 4-10
-----------
This progression is
-------------
A formula would be if you can choose numbers
from 1 to n, then the possible arrangements of 2 numbers
the 1st being smaller than the 2nd
is (n-1) + (n-2) + (n-3) + . . . + 1
This sum is 
check:
for 1 to 10, 
OK
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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
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