Option 3 : 2 × 24P5 × 20!
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Concept: Combinations: The number of ways in which r distinct objects can be selected simultaneously from a group of n distinct objects, is: nCr = \(\rm \frac {n!}{r!(n-r)!}\). Permutations: The number of ways in which r objects can be arranged in n places (without repetition) is: nPr = \(\rm \frac{n!}{(n - r)!}\).
Calculation: There are 26 letters in the English alphabet. If we separate the group (a, some 5 letters, b), we will be left with 19 more letters. These 20 objects (1 group + 19 letters) can be arranged among themselves in 20! ways. Since either a or b can be at the beginning or the end of the group of 7 letters (a, some 5 letters, b), the number of possible arrangements of the group will be 2 × (1P1 × 5P5 × 1P1) = 2 × 5!. Also, each group of 5 letters can be selected from the remaining 24 letters (except a and b) in 24C5 ways. Required total number of ways = (2 × 5! × 24C5) × 20! = 2 × 24P5 × 20!. India’s #1 Learning Platform Start Complete Exam Preparation
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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.
Here is another ways of visualizing the problem. First, we arrange $16$ blue balls and $10$ green balls so that no two of the green balls are consecutive. Then we label the balls. Line up $16$ blue balls, leaving spaces between successive balls and at the ends of the row. There are $17$ such spaces, $15$ between successive blue balls and two at the ends of the row. $$\square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square \color{blue}{\bullet} \square$$ To ensure that no two of the ten green balls are adjacent, we choose $10$ of these $17$ spaces in which to insert a green ball, which can be done in $$\binom{17}{10}$$ ways. Next, we label the balls from left to right. The letters on the green balls are the desired subset of the alphabet in which no two of the selected letters are consecutive. For instance, $$\color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{green}{\bullet} \color{blue}{\bullet} \color{blue}{\bullet}$$ corresponds to the selection $\{A, C, G, J, L, N, R, T, V, X\}$. |
How many ways can two letters be selected from the English alphabet if repetition is allowed?
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