This section covers permutations and combinations. Arranging Objects The number of ways of arranging n unlike objects in a line is n! (pronounced ‘n factorial’). n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1 Example How many different ways can the letters P, Q, R, S be arranged? The answer is 4! = 24. This is because there are four spaces to be filled: _, _, _, _ The first space can be filled by any one of the four letters. The second space can be filled by any of the remaining 3 letters. The third space can be filled by any of the 2 remaining letters and the final space must be filled by the one remaining letter. The total number of possible arrangements is therefore 4 × 3 × 2 × 1 = 4!
n! . Example In how many ways can the letters in the word: STATISTICS be arranged? There are 3 S’s, 2 I’s and 3 T’s in this word, therefore, the number of ways of arranging the letters are: 10!=50 400 Rings and Roundabouts
When clockwise and anti-clockwise arrangements are the same, the number of ways is ½ (n – 1)! Example Ten people go to a party. How many different ways can they be seated? Anti-clockwise and clockwise arrangements are the same. Therefore, the total number of ways is ½ (10-1)! = 181 440 Combinations The number of ways of selecting r objects from n unlike objects is:  Example There are 10 balls in a bag numbered from 1 to 10. Three balls are selected at random. How many different ways are there of selecting the three balls? 10C3 =10!=10 × 9 × 8= 120 Permutations A permutation is an ordered arrangement.
nPr = n! . Example In the Match of the Day’s goal of the month competition, you had to pick the top 3 goals out of 10. Since the order is important, it is the permutation formula which we use. 10P3 =10! = 720 There are therefore 720 different ways of picking the top three goals. Probability The above facts can be used to help solve problems in probability. Example In the National Lottery, 6 numbers are chosen from 49. You win if the 6 balls you pick match the six balls selected by the machine. What is the probability of winning the National Lottery? The number of ways of choosing 6 numbers from 49 is 49C6 = 13 983 816 . Therefore the probability of winning the lottery is 1/13983816 = 0.000 000 071 5 (3sf), which is about a 1 in 14 million chance.
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Misc 2 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? Misc 2 How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together? Number of vowels in EQUATION = E, U, A, I, O = 5 Number of ways vowels can be arranged = 5P5 = 5!/(5 − 5)! = 5!/0! = 5!/1 = 120 Number of consonants in EQUATION = Q, T, N = 3 Number of ways consonants can be arranged = 3P3 = 3!/(3 − 3)! = 3!/0! = 3!/1 = 6 Total number of ways in which vowels & consonants occur together = 2 × (Number of ways vowel arrange) × (Number of ways consonants arrange) = 2 × (120 × 6) = 1440 Answer Hint: Permutations are the different ways in which a collection of items can be arranged. For example:The different ways in which the alphabets A, B and C can be grouped together, taken all at a time, are ABC, ACB, BCA, CBA, CAB, BAC. Note that ABC and CBA are not the same as the order of arrangement is different. The same rule applies while solving any problem in Permutations.The number of ways in which n things can be arranged, taken all at a time, \[{}^n{P_n}{\text{ }} = {\text{ }}n!\], called ‘n factorial.’ Complete step-by-step answer: Total number of letters in “EQUATION” = 8.There are 5 vowels: a, e, i, o, u and 3 consonants : q, t, n.Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects.Then they form 2 groups V(vowels) and C (consonants)We first arrange the 2 groups.The permutations of these 2 objects taken all at a time are counted: ${}^2{P_2} = 2! = 2$waysCorresponding to each of these permutations, Now the group V has 5 elements, they can be arranged in $5! = 120$ ways.Now the group C has 3 elements, they can be arranged in $3! = 6$ ways.Hence by multiplication principle, required number of words = $2! \times 5! \times 3!$the total no of ways = $1440$Therefore, 1440 words with or without meaning, can be formed using all the letters of the word ‘EQUATION’, at a time so that the vowels and consonants occur together. Note: Always keep an eye on the keywords used in the question. The keywords can help you get the answer easily. The keywords like-selection, choose, pick, and combination-indicates that it is a combination question.Keywords like-arrangement, ordered, unique- indicates that it is a permutation question.If keywords are not given, then visualize the scenario presented in the question and then think in terms of combination and arrangement.
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How many words with or without meaning can be formed using all the letters of the word "EQUATION" at a time if vowels and consonants occur together. My answer is $5!3!2!=1440$. Am I right? $\endgroup$ 1 |
How many ways the letters of the word equation can be arranged if the vowels and consonants always occur together?
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