The word ARRANGEMENT has $11$ letters, not all of them distinct. Imagine that they are written on little Scrabble squares. And suppose we have $11$ consecutive slots into which to put these squares. There are $\dbinom{11}{2}$ ways to choose the slots where the two A's will go. For each of these ways, there are $\dbinom{9}{2}$ ways to decide where the two R's will go. For every decision about the A's and R's, there are $\dbinom{7}{2}$ ways to decide where the N's will go. Similarly, there are now $\dbinom{5}{2}$ ways to decide where the E's will go. That leaves $3$ gaps, and $3$ singleton letters, which can be arranged in $3!$ ways, for a total of $$\binom{11}{2}\binom{9}{2}\binom{7}{2}\binom{5}{2}3!.$$
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. Now here we will discuss few question answers on different ways of Letter arrangement which are very common in competitive exams. These Letters arrangement practice question answer session will help you to prepare for your examination. Your math skills is very much needed to solve this kind of problems. Shortcut tricks can also be used to solve different ways of Letter arrangement questions. We try to bring together all types of shortcut methods on Letters arrangement for every topic here in this website. Now you just need to apply those tricks to solve these questions. These questions can be solvable without using any shortcut methods also. Few question on different ways of Letter arrangement will be discuss here. All you need to do is to read the question very carefully and try to solve it by yourself. Answer of this question will be provided along with examples. If you do this problem then check the solution of this question with your answer. If you don’t know how to solve this then also check below. Every page of this section is contain a question on different ways of Letter arrangement with its detail explanation. Next/Previous link will help you to navigate through other questions. Let’s starts the Question Answer session. Example #1 – Different ways of Letter arrangementIn how many different ways can the letters of the word “STUDENT” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (C) How to Solve Total ways of arrangements is, STUDENT = 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. Rough Workspace In how many different ways can the letters of the word “APPLE” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (A) How to Solve Total ways of arrangements is, APPLE = 5! = 5 x 4 x 3 x 2 x 1 = 120. Rough Workspace In how many different ways can the letters of the word “LEARNER” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (D) How to Solve Total ways of arrangements is, LEARNER = 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040. Rough Workspace In how many different ways can the letters of the word “DANGER” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (D) How to Solve Total ways of arrangements is, DANGER = 6! = 6 X 5 X 4 X 3 X 2 X 1 = 720. Rough Workspace In how many different ways can the letters of the word “LAPTOP” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (B) How to Solve Total ways of arrangements is, LAPTOP = 6! = 6 X 5 X 4 X 3 X 2 X 1 = 720. Rough Workspace In how many different ways can the letters of the word “INDIA” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (A) How to Solve Total ways of arrangements is, INDIA = 5! = 5 X 4 X 3 X 2 X 1 = 120. Rough Workspace In how many different ways can the letters of the word “BOLLYWOOD” be arranged? Show Answer Show How to Solve Open Rough Workspace Answer: Option (D) How to Solve Total ways of arrangements is, BOLLYWOOD = 9! = 9 X 8 X 7 X 6 X 5 X 4 X 3 X 2 X 1 = 362880. Rough Workspace How many different ways can be formed by using all the letters of the words “FEBRUARY” so that the vowels always come together? Show Answer Show How to Solve Open Rough Workspace Answer: Option (B) How to Solve The word FEBRUARY contains 8 letters. In this word vowel letters are “EUA”, and we consider it as one single letter. So, letter be arrange as FBRRY (EUA). We can arrange 6 letters as 6P6 or, 6! or, 6 x 5 x 4 x 3 x 2 x 1 therefore, 720 ways. We can also arrange 3 vowel as, or, 3! or, 3 x 2 x 1 therefore, 6 ways. So, required number of arrangements are, or, (720 x 6) therefore, 4320 ways. Rough Workspace How many different ways can be formed by using all the letters of the words “COMPUTER” so that the vowels always come together? Show Answer Show How to Solve Open Rough Workspace Answer: Option (C) How to Solve The word COMPUTER contains 8 letters. In this word vowel letters are “OUE”, and we consider it as one single letter. So, letter be arrange as CMPTR (OUE). We can arrange 6 letters as 6P6 or, 6! or, 6 x 5 x 4 x 3 x 2 x 1 therefore, 720 ways. We can also arrange 3 vowel as, or, 3! or, 3 x 2 x 1 therefore, 6 ways. So, required number of arrangements are, or, (720 x 6) therefore, 4320 ways. Rough Workspace How many different ways can be formed by using all the letters of the words “SISTER” so that the vowels always come together? Show Answer Show How to Solve Open Rough Workspace Answer: Option (A) How to Solve The word SISTER contains 6 letters. In this word vowel letters are “IE”, and we consider it as one single letter. So, letter be arrange as SSTR (IE). We can arrange 5 letters as 5P5 or, 5! or, 5 x 4 x 3 x 2 x 1 therefore, 120 ways. We can also arrange 2 vowel as, or, 2! or, 2×1 therefore, 2 ways. So, required number of arrangements are, or, (120 x 2) therefore, 240 ways. Rough Workspace How many different ways can be formed by using all the letters of the words “KEYBOARD” so that the vowels always come together? Show Answer Show How to Solve Open Rough Workspace Answer: Option (B) How to Solve The word KEYBOARD contains 8 letters. In this word vowel letters are “EOA”, and we consider it as one single letter. So, letter be arrange as KYBRD (EOA). We can arrange 6 letters as 6P6 or, 6! or, 6 x 5 x 4 x 3 x 2 x 1 therefore, 720 ways. We can also arrange 3 vowel as, or, 3! or, 3 x 2 x 1 therefore, 6 ways. So, required number of arrangements are, or, (720 x 6) therefore, 4320 ways. Rough Workspace Answer of this question is provided along with this examples. Scroll down to see the answer. You can take as much time as you need to answer the question. But try to solve this as quicker as you can. Please visit this page to get updates on different ways of Letter arrangement. And for more visit Math Shortcut Tricks. You can also like our facebook page to get updates.Few questions on some other topics are also provided here. Click on the Next and Previous link to get few more questions and answers. If you have any question regarding different ways of Letter arrangement, then please do comment on below section.You can also send us message on facebook. |
What is the correct arrangement of these letters answer
Postagens relacionadas
direito autoral © 2024 estudarpara Inc.
