0
Want this question answered?
Be notified when an answer is posted
Add your answer:
Earn +20 pts
How many ways can 9 performances at a piano recital be ordered?
9x8x7x6x5x4x3x2x1 = 362880
Exclamation point in a math equation?
The exclamation point is called a factorial. ex. 5!=5x4x3x2x1 10!=10x9x8x7x6x5x4x3x2x1 etc.
How many different permutations are in the word ASSESSMENT?
There are 75600 permutations.
How many times can you rearrange the order of 10 books?
You can arrange them as many times as you want. However, there are a limited number of orders to put them in, 10! (That's ten factorial, not an exclamation point.) 10x9x8x7x6x5x4x3x2x1 ways, or 3628800 ways.
How many ways can 10 students line up for lunch?
for the first student in the line there are 10 choices, then for the second 9 choices left, for the third 8 choices left and so on... So it's 10x9x8x7x6x5x4x3x2x1 = 3628800
In how many ways can 10 people line up at a ticket window of a cinema hall?
If you imagine there are 10 spots on the floor in a line where people can stand, then the first person has 10 choices of where to stand. The second person has 9 choices since the first person is already on one of the spots. The third person has 8 choices. And so on until the last person has no choice. We can now mulitply the numbers together. 10x9x8x7x6x5x4x3x2x1. This is called a factorial (where you multiply all of the consecutive numbers together from 1) and it is written as 10!. This means there are 3,628,800 ways of arranging the 10 people
What are the combinations for 1234567890?
There are over 3.6 billion possible combinations for a 10-digit number like 1234567890. This number can be rearranged in various ways, but each combination would still consist of the same set of digits (1, 2, 3, 4, 5, 6, 7, 8, 9, 0).
How many different combinations can you make out of 5 different things?
5! = 120 ! means factorial. A factorial is the product of of the positive integers and equals the number of different combinations of a number. A factorial can be work out quite simply. Take the number 5. 5! = 5x4x3x2x1 = 120 So simply place the number you are trying to find out the combinations for first and then times it by all the numbers below. Some more examples would be: 8! = 8x7x6x5x4x3x2x1 = 4320 3! = 3x2x1 = 6 10! = 10x9x8x7x6x5x4x3x2x1 = 3,628,800 6! = 6x5x4x3x2x1 = 720 * * * * * An interesting introduction on factorials but totally misses the point of the question. A factorial generates permutations, not combinations! For combinations, abc is the same as acb, cab, bac, etc. The number of combinations of that you can make out of 5 things *including the null combination - ie nothing) is 25 = 32.
What is the probability that 13 rolls of a fair die will show three threes?
Not an easy question. Here's a long but logical answer. The chance of not getting a 3 on any throw is 5/6, so the chance of not getting any threes in the first 10 throws is (5/6)^10. Now, having reached that point, the chance of throwing three threes in the last three throws is 1/6 x 1/6 x1/6 or 1/216. So the probability of throwing three threes in any chosen three throws is (5/6)^10 x 1/216. This should seem logical because the probability of throwing three threes in the first three throws followed by ten failures to throw threes, should be the same as just getting threes in the last three throws. The probability is the same for getting threes in only the second, fourth and sixth throw, or any other combination of three throws. Now, there are a set number of combinations of 13 things taken in groups of three. This is given by the formula for C(13,3) = (13x12x11x10x9x8x7x6x5x4)/((10x9x8x7x6x5x4x3x2x1)x(3x2x1)). That cancels down to (13x12x11)/(3x2x1), and equals 286. So, there are this many ways you can pick out three throws which you want to be the threes, and the probability of them actually being threes is the first calculation we did. The answer to the question is 286 x (5/6)^10 x 1/216 which is 0.2138453, about.
What is the formula for permutations?
The number of permutations of r objects taken from n distinct objects is nPr = n!/(n-r)! n! = 1*2*3*... *(n-1)*n In many cases, one can do permutation problems without the "formula." Here is a concrete example, then a more abstract version to help see this. The purpose of pointing this out is not only to make it easier to do this type of problem, but to help understand and remember the formula. Say we have 10 distinct (we assume distinct from now on) objects and we want to pick 3 objects from the 10. The order of choice is important. Since there are 10 objects, there are 10 choices when we pick the first one. Now there are 9 left since one is picked so there are 9 choices for the next object. Similarly, there are 8 choices for the third object. The multiplication rule tells us there are 10x9x8=720 ways to pick these three objects. Now say we have n distinct objects and we want to pick r of them. We have n choices for the first, n-1 choices for the second and n-3 choices for the third. If n=10 as above and r=3, we saw this was 10x9x8 but in the more general case it is n x (n-1) x(n-2). But how do we know where to stop? We want to pick r items. So in the case of r=3, the last number in our product was 8. This is (10-3)+1. So in general the last number in the product is (n-r)+1 The general pattern therefore for picking r objects taken from n distinct objects is n(n-1)(n-2)(n-3)...(n-r+1). This is the same as n!/(n-r)! The denominator in the formula simply allows us to write n! which is 1*2*3*... *(n-1)*n and cancel out all the numbers in the product that are less than n-r+1. For example. Using the same example if we pick 3 items from 10 we saw that was 10x9x8. Using the formula we have 10!/(10-3)! or 10!/7! which is (10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1) and after cancelling the (7x6x5x4x3x2x1), which occurs in the products in both the numerator and the denominator, we are left with 10x9x8=720 which we know to be the answer.
How do you write an algorithm to find the number of permutations or combinations?
Permutations and combinations are two separate things. Although we often use them interchangeably in English, we need a more precise definition in mathematics, such that ABC and CBA are regarded as being two different permutations of the same combination. In other words, the order of the elements is important in a permutation but is completely irrelevant in a combination.First we have to define what it means to create a combination or permutation. Typically we have a set from which we must make a subset. The number of elements in the set is typically defined using the variable n while the number of elements in the subset is r. Thus we can formally define a permutation mathematically using the function P(n,r) and a combination as C(n,r).We must also consider whether elements may be repeated within a combination or a permutation. For instance, when selecting numbers for a lottery, no number may be repeated in any combination but in a 4-digit combination lock, any digit may be repeated in a permutation.Note that a combination lock is really a permutation lock in mathematics and is the perverse way of remembering the mathematical difference between a combination and a permutation.Thus we have 4 possible variations to cater for. In order of difficulty, they are:Permutations with repetitionPermutations without repetitionCombinations without repetitionCombinations with repetitionLet's deal with them one at a time.1. Permutations with repetitionTo calculate P(n,r) with repetitions, for every selection, r, there are always n possibilities, thus we have n^r permutations.In a 3-digit combination lock, each digit has ten possibilities, 0 through 9, so there are 10^3=10x10x10=1000 permutations. This stands to reason because the permutations form all of the numeric values from 000 through to 999, which is 1000 different values.In C, we can write this function as:unsigned long long permutations (unsigned long long n, unsigned long long r) {return pow(n,r); /* use standard library function */}2. Permutations without repetitionTo calculate P(n,r) without repetitions we must reduce the set by one element each time we make a selection. If we go back to our 3-digit combination lock, we have 10 choices for the first digit which leaves 9 choices for the next and 8 for the next. So instead of 10x10x10=1000 permutations we only have 10x9x8=720 permutations.Although fairly simple to work out in this case, we need a formula that is generalised to cater for all cases, just as n^r works for all permutations with repetitions.We can see that 10x9x8 is the initial product of 10! (factorial 10) which is 10x9x8x7x6x5x4x3x2x1. So we need a formula that ignores everything after the 8. The portion after the 8 is 7x6x5x4x3x2x1 which is 7! and we can calculate that from (n-r)!=(10-3)!=7!Having determined the portion we need to ignore, the rules of multiplication and division state that if we multiply by x and subsequently divide by x, then the two x's must cancel each other out. Thus we get:10!/7!=(10x9x8x7!)/7!=10x9x8=720Using formal notation, P(n,r) without repetition is therefore n!/(n-r)!In C, we must first write a function to calculate factorials:unsigned long long factorial (unsigned long long n) {return (n>1)?factorial(n-1):1; /* recursive function */}With that in place, we can now write a function to handle permutations without repetitions:unsigned long long permutations_norep (unsigned long long n, unsigned long long r) {return factorial(n)/factorial(n-r);}3. Combinations without repetitionC(n,r) without repetition is simply an extension of P(n,r) without repetition. Every combination of r has r! permutations, so if we divide P(n,r) by r! we will get C(n,r). Expressing this formally, C(n,r) without repetition is n!/((n-r)!r!)Going back to our 3-digits from 10, there are 10!/((10-3)!3!)=10!/(7!3!)=(10x9x8x7!)/(7!3!)=(10x9x8)/3!=720/6=120 combinations without repetition.Using the factorial function shown above, we can write a C function to handle combinations without repetition:unsigned long long combinations_norep (unsigned long long n, unsigned long long r) {return factorial(n)/(factorial(n-r)*factorial(r));}4. Combinations with repetitionCombinations with repetition is the hardest concept to wrap your head around.Going back to our 3-digits from 10, let's begin enumerating all the combinations so we can verify the answer at the end. We start by enumerating all those that combinations that begin with a 0:000, 001, 002, 003, 004, 005, 006, 007, 008, 009011, 012, 013, 014, 015, 016, 017, 018, 019022, 023, 024, 025, 026, 027, 028, 029033, 034, 035, 036, 037, 038, 039044, 045, 046, 047, 048, 049055, 056, 057, 058, 059066, 067, 068, 069077, 078, 079088, 089099Note that there is no 010 because it is a permutation of 001. Similarly with 021 which is a permutation of 012. As a result of this, each row has one less combination than the one above. Thus there are 10+9+8+7+6+5+4+3+2+1=55 combinations.If we now enumerate all those that begin with a 1, we see a similar pattern emerges:111, 112, 113, 114, 115, 116, 117, 118, 119122, 123, 124, 125, 126, 127, 128, 129133, 134, 135, 136, 137, 138, 139144, 145, 146, 147, 148, 149155, 156, 157, 158, 159166, 167, 168, 169177, 178, 179188, 189199This time we have 9+8+7+6+5+4+3+2+1=45 combinations.Following the same logic, the next section must have 8+7+6+5+4+3+2+1=36 combinations, followed by 28, 21, 15, 10, 6, 3 and finally 1. Thus there are 220 combinations in total.The formula to work this out is quite complex, however it becomes simpler when we look at the problem in a different way. Suppose we have 10 boxes and each box holds at least 3 of the same digit. We can number these boxes 0 through 9 according to those digits. Let us also suppose that we can only move in one direction, from box 0 to box 9, and we must stop at every box along the way. This means we must make 9 transitions from one box to the next.While we along the row, we carry a tray with 3 slots. Whenever we stop at a box (including box 0 where we start from) we can either pick a number from the box or we can move onto the next box. if we pick a number, we place it in the first slot. We can then pick another or we can move on. When we have filled all the slots, we simply move on until we reach box 9. If we reach box 9 and still have slots available, we must pick as many 9s as we need to fill the remaining slots.It probably sounds far more complex than it really is. By imagining a selection being done this way we can create a convenient binary notation. For instance, if we say that 1 means pick a number and 0 means move onto the next box, the sequence 101010000000 would tell us we selected the combination 123 while the sequence 000000000111 tells us we selected 999. Every combination is therefore reduced to 12-bit value containing exactly three 1s and nine 0s, and it is these specific combinations we are actually looking for.C(n,r) with repetition is formally expressed as (r+n-1)!/(r!(n-1)!)If we plug in the actual numbers we find:=(3+10-1)!/(3!(10-1)!)=12!/(3!9!)=(12x11x10x9!)/(3!9!)=(12x11x10)/3!=1320/(3x2x1)=1320/6=220 combinations with repetition.This type of problem might be expressed in other ways. For example, how many different ways can we fill a box with 100 sweets from 30 different sweets. C(n,r) is C(30,100) thus we find:=(100+30-1)!/(100!(30-1)!)=129!/(100!29!)=(129x128x127x...x101x100!)/(100!29!)=(129x128x127x...x101)/29!=5.3302324527079900778691094496787e+59/8,841,761,993,739,701,954,543,616,000,000=60,284,731,216,266,553,294,577,246,880 combinations with repetition.In C we can use the following function in conjunction with the factorial function shown earlier:unsigned long long combinations (unsigned long long n, unsigned long long r) {return factorial(n+r-1)/(factorial(r)*factorial(n-1));}We might also have similar problems with an additional restriction. For instance, we might be asked to select 100 sweets from 30 different sweets selecting at least 1 of each type. This reduces the number of slots to 100-30=70 but we have the same number of transitions, so we get:=(70+30-1)!/(100!(30-1)!)=99!/(70!29!)=(99x98x97x...x71x70!)/(70!29!)=(99x98x97x...x71)/29!=7.7910971370578048745872324992773e+55/8,841,761,993,739,701,954,543,616,000,000=8,811,701,946,483,283,447,189,128 combinations with repetition.To accommodate this caveat, we can use the following function instead:unsigned long long combinations2 (unsigned long long n, unsigned long long r) {return factorial(n-1)/(factorial(r)*factorial(n-1));}
