Mathematics Exam  >  Mathematics Questions  >  Five digil number divisible by 9 are to be fo... Start Learning for Free
Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal to
  • a)
    216
  • b)
    214
  • c)
    312
  • d)
    200
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Five digil number divisible by 9 are to be formed by using the digit 0...
 If the sum of digit is dvisible by 9 then number is divisible by 9.
So, (0,7), (2,5) or (3,4)
will not be taken in formation of number.
So no. of ways 5!+2 (4 (4 !))
= 120 + 2(96)= 312
View all questions of this test
Most Upvoted Answer
Five digil number divisible by 9 are to be formed by using the digit 0...
Problem:
We need to find the total number of 5-digit numbers that can be formed using the digits 0, 1, 2, 3, 4, 7, and 8 without repetition, where each number is divisible by 9.

Solution:
To solve this problem, we can follow these steps:

Step 1: Identify the divisibility rule for 9:
A number is divisible by 9 if the sum of its digits is divisible by 9. So, we need to find the numbers that have a digit sum of 9.

Step 2: Determine the number of choices for each digit:
Since we want to form 5-digit numbers, we need to determine the number of choices we have for each digit. We can do this by counting the number of available digits for each position.

For the first digit, we have 7 choices (0, 1, 2, 3, 4, 7, 8).
For the second digit, we have 6 choices (any digit except the one already chosen).
For the third digit, we have 5 choices.
For the fourth digit, we have 4 choices.
For the fifth digit, we have 3 choices.

Therefore, the total number of choices for all five digits is 7 x 6 x 5 x 4 x 3 = 5040.

Step 3: Determine the number of valid combinations:
We need to find the number of combinations that have a digit sum of 9. To do this, we can use the concept of combinations.

Since the digits cannot be repeated, we need to find the number of combinations of 5 digits taken from the available choices (7 digits).

The formula for combinations is nCr = n! / (r!(n-r)!), where n is the total number of options and r is the number of choices.

Using this formula, we can calculate the number of combinations:

7C5 = 7! / (5!(7-5)!) = 7! / (5!2!) = (7 x 6 x 5 x 4 x 3 x 2 x 1) / ((5 x 4 x 3 x 2 x 1) x (2 x 1)) = 7 x 6 / 2 = 42.

So, there are 42 valid combinations.

Step 4: Determine the total number of valid numbers:
Now that we have the number of valid combinations, we need to find the total number of valid numbers. Since we have 5 digits, the number of valid numbers is equal to the number of valid combinations multiplied by the number of choices for each digit.

Total number of valid numbers = 42 x 5040 = 211,680.

Therefore, the correct answer is option A) 216.
Explore Courses for Mathematics exam
Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer?
Question Description
Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? for Mathematics 2024 is part of Mathematics preparation. The Question and answers have been prepared according to the Mathematics exam syllabus. Information about Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for Mathematics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer?.
Solutions for Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Mathematics. Download more important topics, notes, lectures and mock test series for Mathematics Exam by signing up for free.
Here you can find the meaning of Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer?, a detailed solution for Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? has been provided alongside types of Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice Five digil number divisible by 9 are to be formed by using the digit 0,1,2,3,4,7,8 (without repetition). The total number of such numbers is equal toa)216b)214c)312d)200Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice Mathematics tests.
Explore Courses for Mathematics exam
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev