    # How many three digit numbers can be formed using the digits 5 6 7 and 8 without repetition?

 \$\begingroup\$ Given the numbers \$1\$, \$5\$, \$7\$, \$8\$, and \$9\$, how many \$3\$-digit numbers larger than \$700\$ can be formed if repetition is not allowed? The answer is \$36\$. I want a detailed explanation please of how we get this answer? \$\endgroup\$ 0  Prev Question 22 Exercise 1(A) Next Answer: Solution : All the possible three digits number using 3,6 and 8 only; if the repetition of digits is not allowed , can be 368, 386, 638, 836, 863 Video transcript "Hello children, welcome to lido homework. So now we are doing question number 19. So what this question asked us is to write all possible three digit numbers using the digits three six and eight only if reputation of digits is not allowed. So repetition is not allowed. So let's start writing. So first of all, let's write them in order. So it will be 368. This is our first digit, right? Let's move forward. Let's interchange 3 and 6. It will become 6. 638 on moving forward let's interchange three and eight now so it will become 683. Let's move forward now, let's take let's take it first. So it will be 863 or 836 and the last possible is 3/8. 6 so these are all your possible numbers of digits with the digits 368 when reputation is not allowed. Now, if reparations allowed this would go on to be an infinite number because we could never stop we could keep on writing six six six six three three three three and not stop so we cannot list out all of them if repetition of diseases allow. So this is your answer to this question. Thank you so much. " Was This helpful?   India's Super Teachers for all govt. exams Under One Roof Enroll For Free Now Concept: Fundamental Principle of Multiplication: Let us suppose there are two tasks A and B such that the task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to complete the task A and B in succession respectively is given by: m × n ways. Fundamental Principle of Addition: Let us suppose there are two tasks A and B such that the task A can be done in m different ways and task B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by: (m + n) ways. Calculation: Here, we have to find how many 3-digit numbers can be formed without using the digits 0, 2, 3, 4, 5 and 6. i.e we have to find how many 3-digit numbers can be formed using the digits 1, 7, 8, 9. Clearly, repetition of digits is allowed. The number of ways to fill unit's digit = 4 The number of ways to fill ten's digit = 4 The number of ways to fill hundred's digit = 4 ∴ Total number of required numbers = 4 × 4 × 4 = 64 India’s #1 Learning Platform Start Complete Exam Preparation Daily Live MasterClasses Practice Question Bank Mock Tests & Quizzes Get Started for Free Download App Trusted by 3.4 Crore+ Students Text Solution3456Answer : DSolution : 357, 375, 537, 573, 735, 753. Therefore, '6' three-digit numbers can be formed.
Hence, the correct option is (d). Answer:24Step-by-step explanation:let these letters stand for a number. ABCSince we have 4 numbers, A can be filled 4 times. B can be filled 3 timesC can be filled 2 times.Hence the total number is 4*3*2=24 