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Given: 4 digit numbers to be formed using the digits 1, 2, 3, 4, .....8 such that no two consecutive digits are the same Concept used: Number system Calculation: Here, no two consecutive digits can be the same. ⇒The thousands place can be filled by any of the 8 digits (1, 2, 3, 4, .....8) ⇒ Repetition is allowed here. The only restriction is that no two consecutive digits can be the same. The digit we placed in the thousands place cannot be used at the hundreds place. Hence thousands place can be filled by any of the 8 digits. ⇒ Similarly, tens place and unit place can be filled by any of the 7 digits Hence, the required count of 4 digit numbers that can be formed using the digits 1, 2, 3, 4 .....8 such that no two consecutive digits are the same ⇒ 8 × 73 ∴The required number of ways is 8 × 73 India’s #1 Learning Platform Start Complete Exam Preparation
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Numbers that follow each other continuously in the order from smallest to largest are called consecutive numbers. For example: 1, 2, 3, 4, 5, 6, and so on are consecutive numbers. Consecutive Even Numbers:Even numbers are numbers that end with 0, 2, 4, 6 or 8. The examples of consecutive even numbers are:0, 2, 4, 6, 8, 10, 12, …. Consecutive Odd Numbers:Odd numbers are numbers that end with 1, 3, 5, 7 or 9. The examples of consecutive odd numbers are:1, 3, 5, 7, 9, 11, 13, 15, …. Consecutive Even and Odd Integers:We can also have consecutive even and odd integers. Example: Consecutive Even Integers: – 8, –6, –4, –2, 0, 2, 4, 6, ….. Example: Consecutive Odd Integers: –9, –7, –5, –3, –1, 1, 3, 5, 7, ….The term consecutive numbers is often used to frame word problems. Example: The sum of two consecutive numbers is 55. What are the numbers? Here, let the first number be a. Since the numbers are consecutive, the other number will be a + 1 We now form an equation as per the given information: Sum of the numbers = 55 = a + a + 1 We should choose the numbers such that their sum is 55. 26 is nearly half of 55. Let the two number be 26 and 27; 26 + 27 = 53 ✘ So, the two numbers are not 26 and 27. Let us choose the next number, 28. So, the two numbers are 27 and 28. 27 + 28 = 55 ✓Therefore, the numbers are 27 and 28. Example: The product of two consecutive odd numbers is 143. What are the numbers? We should choose the numbers whose product is nearly 143. We know that 12 × 12 = 144 But, 12 is an even number. Consecutive odd numbers near 12 are 11 and 13. Let us find their product; 11 × 13 = 143 ✓Therefore, the numbers are 11 and 13.
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My approach is to solve this problem using recursion. Let T(n) be the number of such n digit integers such that no 3 consecutive digits are 8. Now there are a few cases: Case 1: Last digit is not 8. So we are left with dealing with the remaining n-1 digits where the last digit can take any value except 8 i.e. the number of such numbers are 9T(n-1) since last digit can be anything but 8 and we have to apply recursion on the remaining n-1 digits. Case 2: Last digit is 8. Now, here it gets a bit tricky. If the last digit is 8, then again for the 2nd last digit, we have two options, it is either 8 or it is not 8. If it is not 8, we have no problem , the number of cases will be 9T(n-2) since the last digit is 8, and the 2nd last is anything but 8, and we recurse on the previous n-2 digits in the integer.If the 2nd last digit is 8, then again, we must choose the 3rd last digit as not 8, so to avoid the subsequence 888. This can be done in 9T(n-3) ways as only the 3rd last digit should not be 8, and the remaining digits are already 8, so 9T(n-3), similarly recursing on the remaining n-3 digits. So according to me number of such n digit integers such that no three digits are 8 can be solved by solving the recursion T(n) = 9T(n-3)+9T(n-2)+9T(n-1) The base cases are T(1) = 9, T(2)=90, T(3) = 899. Is this solution correct? Thanks in advance. Answer Hint: We are to find the number of n-digit numbers where no two consecutive digits are the same. We know that a number system has a total of 10 digits (0 to 9). Then, by using permutation and combination, we will check the number of ways each place can be filled. And depending on this, we will see the nth place we can fill in how many ways. By doing so, we will get the final output. Complete step-by-step answer: As we know that, there are a total 10 digits i.e. 0 to 9 in the number system.Also, given that, there are n-digits in the given number.So, here the first place can have any of the 9 digits except 0.This means that the first place can be filled in 9 ways.Also, the second place can have any of the 9 digits except the digit which is at the first place.This means that the second place can be filled in 9 ways.Similarly the third, fourth,…,nth place can have any of the 9 digits.This means that all the places can be filled in 9 ways.Thus, the total number of ways to fill the nth place\[ = 9 \times 9 \times ........ \times 9 \times 9\] (n times)\[ = {9^n}\]Hence, number of n-digit numbers, with no two consecutive digits same is given by \[{9^n}\]So, the correct answer is “Option 3”. Note: Another Method: Given that, no two consecutive digits are the same.This means that there are n numbers starting from 1 and not 0 in the first place.Let there be 1 to 9 numbers.So, the number of ways to fill the first place of the n-digit number = 9 ways.Now, here we are given that no two consecutive numbers are the same.So, if we keep number 2 in the first place, then we can write numbers 0, 1,…, 9 from second place onwards.Thus, the second place of the n-digit number can be filled in 9 ways.Hence, the total number of ways to fill first 2-places of n-digit number \[ = 9 \times 9\]\[ = 81\]\[ = {9^2}\]Similarly, total number of ways to fill n-places of n-digit number is\[ = {9^n}\] |
Find the number of ways of forming 8 digit numbers such that no two consecutive digits are same.
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