Expert II Permutations And Combinations (12), Expert II Permutations et combinaisons (12) In 30-2 and 30-1 Flashcards
(23 cards)
What is factorial?
Qu’est-ce que la factorielle ?
A multiplication of the given number by all natural numbers below it.
Multiplication du nombre donné par tous les nombres naturels inférieurs.
How do you represent a factorial?
Comment représente-t-on une factorielle ?
!
What is 6! in expanded notation?
Quelle est la valeur de 6! en notation élargie ?
6 x 5 x 4 x 3 x 2 x 1
What is 6! ? Show your work.
Qu’est-ce que 6! ? Montrez votre travail.
6 x 5 x 4 x 3 x 2 x 1 = 30 x 12 x 2
= 30 x 24
= 720
Use a calculator to compute 6! .
Utilisez une calculatrice pour calculer 6! .
On a graphing calculator to get the factorial symbol press the following:
Math -> -> -> 4
6! =720
Sur une calculatrice graphique, pour obtenir le symbole factoriel, appuyez sur les touches suivantes :
Math -> -> -> 4
6 ! =720
How can you show any number factorial?
Comment peut-on montrer la factorielle d’un nombre quelconque ?
n! is the most common way
n ! est la méthode la plus courante
What does (n-1)! signify?
Que signifie (n-1)! ?
Subtract 1 from the number you have decided n is equal to at the moment. Then compute the factorial of one less than that number.
If n = 7 then the expression is equal to 6! = 720
Soustrayez 1 du nombre auquel vous avez décidé que n est égal à ce moment-là. Calculez ensuite la factorielle d’un nombre inférieur à ce nombre.
Si n = 7, l’expression est égale à 6 ! = 720
How can you cancel out factorials in expressions that contain n! and n minus any natural number factorialed?
Comment annuler les factoriels dans les expressions contenant n ! et n moins n’importe quel nombre naturel factorisé ?
Take the larger number which is n! and change it to n(n-1)! Keep subtracting one to find more factors until the last one matches the other part of that original fraction.
Then cancel out the factorials that match perfectly.
Prenez le plus grand nombre qui est n ! et changez-le en n(n-1) ! Continuez à soustraire un facteur pour en trouver d’autres jusqu’à ce que le dernier facteur corresponde à l’autre partie de la fraction originale.
Annulez ensuite les factoriels qui correspondent parfaitement.
How many permutations can you make on three tiles selected one at a time without replacement? Assume each tile is unique and all tiles must be selected.
Combien de permutations pouvez-vous faire avec trois tuiles sélectionnées une à la fois sans remplacement ? Supposez que chaque tuile est unique et que toutes les tuiles doivent être sélectionnées.
3! = 3 x 2 x 1
There are three options for the first round so n = number of options for first round = 3
There are three rounds so r = 3
The formula for a permutation is n! / (n - r) !
Since we are sampling until the end and all tiles must be used the denominator is equal to (3-3)! = 0! = 1 and this does not affect the result.
So in this case you can use n! to find the answer since there are no repeated tiles (they are all unique) and since you are using all tiles (the number of rounds equals the number of options for the first round).
3 ! = 3 x 2 x 1
Il y a trois options pour le premier tour, donc n = nombre d’options pour le premier tour = 3
Il y a trois tours, donc r = 3
La formule pour une permutation est n ! / (n - r) !
Puisque nous échantillonnons jusqu’à la fin et que toutes les tuiles doivent être utilisées, le dénominateur est égal à (3-3) ! = 0 ! = 1 et cela n’affecte pas le résultat.
Dans ce cas, vous pouvez donc utiliser n ! pour trouver la réponse puisqu’il n’y a pas de tuiles répétées (elles sont toutes uniques) et que vous utilisez toutes les tuiles (le nombre de tours est égal au nombre d’options pour le premier tour).
How many permutations can you make using exactly three letters from a seven letter tiles? Assume all letters are unique.
Combien de permutations pouvez-vous faire en utilisant exactement trois lettres d’une série de sept lettres ? Supposez que toutes les lettres sont uniques.
Permutations = n! / (n-r)!
n = number of options for first round = 7
r = rounds = 3
Permutations = 7! / (7 - 3)!
= 7! / 4!
= 7 x 6 x 5
It is important to know every step listed here for future understanding of all types of permutations.
In addition, know how to input this in the calculator:
nPr
Most importantly, recognize that all tiles were unique. If they were not you have to further divide the answer to ensure that you have less responses since GOOD is the same as GOOD, even when the O’s switch places. In that case you need to divide by the factorial of the quantity of that doubled letter. You can do this again for another type of doubled letter. This will also work for combinations later.
Permutations = n ! / (n-r) !
n = nombre d’options pour le premier tour = 7
r = tours = 3
Permutations = 7 ! / (7 - 3) !
= 7 ! / 4 !
= 7 x 6 x 5
Il est important de connaître toutes les étapes énumérées ici pour mieux comprendre tous les types de permutations.
De plus, vous devez savoir comment entrer ces données dans la calculatrice :
nPr
Le plus important est de reconnaître que toutes les tuiles sont uniques. Si ce n’est pas le cas, vous devez diviser la réponse pour vous assurer d’avoir moins de réponses, puisque BON est identique à BON, même si les O changent de place. Dans ce cas, vous devez diviser par la factorielle de la quantité de la lettre doublée. Vous pouvez répéter cette opération pour un autre type de lettre doublée. Cela fonctionnera également pour les combinaisons plus tard.
What is a permutation?
Qu’est-ce qu’une permutation ?
Where order matters
DOG is different from GOD so they make two options rather than one.
Lock combinations are actually permutations since order matters.
Passwords are permutations.
Ingredients in a cake are combinations but layers of a cake could be considered permutations.
Quand l’ordre compte
DOG est différent de GOD, ce qui fait qu’il y a deux options au lieu d’une.
Les combinaisons de serrures sont en fait des permutations puisque l’ordre est important.
Les mots de passe sont des permutations.
Les ingrédients d’un gâteau sont des combinaisons, mais les couches d’un gâteau peuvent être considérées comme des permutations.
What does n signify in the context of permutations and combinations?
Que signifie n dans le contexte des permutations et des combinaisons ?
n is the number of options for the first round, where usually this number will decrease by 1 in subsequent rounds
n est le nombre d’options pour le premier tour, ce nombre diminuant généralement de 1 pour les tours suivants.
What does r signify in the context of permutations and combinations?
Que signifie r dans le contexte des permutations et des combinaisons ?
r is the number of rounds
For example, the amount of times you will roll the die, or the length of the word you are spelling
r est le nombre de tours
Par exemple, le nombre de fois où vous lancerez le dé, ou la longueur du mot que vous épellez.
When you answer a permutation or combination, what is the name of the solution?
Lorsque vous répondez à une permutation ou à une combinaison, quel est le nom de la solution ?
The number of outcomes
Le nombre de résultats
When you are creating permutations and combinations and there are some options that are identical, what do you need to do in order to properly count the outcomes?
Lorsque vous créez des permutations et des combinaisons et que certaines options sont identiques, que devez-vous faire pour compter correctement les résultats ?
Find the outcomes by doing the regular permutation or combination first, but then remember to divide by the factorial of the amount of that identical item. If there are three Ts, then divide by 3! =3 x 2 x 1 = 6. Do this for each identical type.
Trouvez les résultats en effectuant d’abord la permutation ou la combinaison normale, mais n’oubliez pas de diviser par la factorielle de la quantité de l’élément identique. S’il y a trois T, divisez par 3! =3 x 2 x 1 = 6. Procédez ainsi pour chaque type d’article identique.
How do you complete the path questions?
Keep adding the two numbers from where you could have come from to create the next number.
In the context of the photo, add the number below to the number to the right. Complete this until you get to the destination and the final number you write is the total amount of possible pathways to get to the destination.
For the first rectangle you can use 5C2 or 5C3 to get 10 pathways.
For the 2 by 2 square you can use 4C2 = 6 pathways.
For the final rectangle you can use 2C1 = 2 pathways.
To get the total pathways for all three rectangles you can multiply them together (5C2)(4C2)(2C1) = 1062 = 120 pathways.
The formula is (length + width)C(length or width). You can pick either the length or the width and the answer will be the same. Then multiply these together to work with more than one rectangle.
If there are gaps in the rectangles then do not use the combination notation and just count the pathways manually.
Example problems with repetitions
If two tiles look the same, such as in the tiles Z, O, and O, then you need to account for the fact that Z O O is the same outcome as Z O O even if the O tiles got switched, and all other cases like this.
The way this is handled is to ensure to divide the total outcomes without the idea of repetitions by all the outcomes where that tile could have been replaced with one identical to it.
In short, count how many times O occurs and divide by the factorial of that number. So divide by 2! if O occurs twice. Divide by 3! if you see a letter that is tripled. For every time that you see identical tiles, divide by the factorial of that number of times that tile occurs.
Example problems with restrictions
Restrictions are when a place can only be occupied by some of the tiles.
Write lines to show all the places, and then fill in the options for the places that have the greatest restrictions first. Then realize that the remaining places have one less option than before if one of the tiles has already been picked for a restricted zone. That is why we deal with the restrictions first, since they influence what can be placed in the remaining spots.
Example problems with groups
When certain tiles must be placed together, this can be best achieved by thinking of the little group as one tile, then working on counting the options as if that was all you have, perhaps a shorter word for example. Then, calculate all the ways that one tile could be assuming that they need to be together but that they have options on their order within that small group. Then multiply all the calculations together.
More example problems with groups
Keep in mind that each adult is different from the next here and that we do not have repetitions. Maybe thinking of A1, A2 etc can help you see this.
Binomial Theorem (30-1 only)
statement that for any positive integer n, the nth power of the sum of two numbers a and b, so (a+b)n may be expressed as the sum of n + 1 terms of the form
(nCr)(an-r)(br)
in the sequence of terms, the index r takes on the successive values 0, 1, 2,…, n. The coefficients, called the binomial coefficients, are defined by the formula
nCr
The coefficients may also be found in the array often called Pascal’s triangle
The theorem is useful in algebra as well as for determining permutations and combinations and probabilities. For positive integer exponents, n, the theorem was known to Islamic and Chinese mathematicians of the late medieval period. Al-Karajī calculated Pascal’s triangle about 1000 ce, and Jia Xian in the mid-11th century calculated Pascal’s triangle up to n = 6. Isaac Newton discovered about 1665 and later stated, in 1676, without proof, the general form of the theorem (for any real number n), and a proof by John Colson was published in 1736. The theorem can be generalized to include complex exponents for n, and this was first proved by Niels Henrik Abel in the early 19th century.
What is (a+b)²?
(a+b)² = a² + 2ab + b²
Notice this is the binomial theorem, looking at Pascal’s triangle, we can find 1, 2, 1 in the third row. Use the row that is one more than the exponent number.
n = exponent, so 2 in this case
r = 0, then 1, then 2
(nCr)(an-r)(br) …. = (2C0)(a2-0)(b0) + (2C1)(a2-1)(b1) + (2C2)(a2-2)(b2) = a² + 2ab + b²
It is a lot simpler to just use Pascal’s triangle, since the formula is long and requires combination notation. However, in a problem with an exponent that is much higher, you could program a computer to make the calculations using the combination notation.
What is (a+b)³?
Use row 4 of Pascal’s triangle to find the coefficients.
You probably have true three memorized, so use it to find the next row:
Row three = 1 2 1
Row four = 1 3 3 1
Then decrease the exponent on a by one as you increase the exponent on b by 1.
1a³b⁰ + 3a²b¹ + 3a¹b² + 1a⁰b³
= a³+ 3a²b + 3ab² + b³
