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A five digit number divisible by 3 is to be formed using the numerals 0,1,2,3,4 and 5 without repetition. The total number of ways in which this can be done is:
  • a)
    220
  • b)
    600
  • c)
    240
  • d)
    None of the above
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A five digit number divisible by 3 is to be formed using the numerals ...
The five digit number will either have 0 or it will not have 0.
Case i:
0 is included.
The five digits must be 0, 1, 2, 4, 5 (as the number has to be divisible by 3)
The number of five -digit numbers = 4 × 4 × 3 × 2 × 1 = 96
Case ii:
0 is excluded.
In this case, the five digits should be 1, 2, 3, 4 and 5.
Number of five digit numbers = 5 × 4 × 3 × 2 × 1 = 120
∴ Required count = 120 + 96 = 216 ; Hence,option D is the answer
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Most Upvoted Answer
A five digit number divisible by 3 is to be formed using the numerals ...
To form a five-digit number divisible by 3 using the numerals 0, 1, 2, 3, 4, and 5 without repetition, we need to follow certain rules. Let's break down the problem into steps and solve it systematically.

Step 1: Find the sum of all the digits (0+1+2+3+4+5=15).

Step 2: Determine the possible remainders when the sum of digits is divided by 3. In this case, the sum is 15, so the possible remainders are 0, 1, and 2.

Step 3: Distribute the digits among the five places (thousands, hundreds, tens, units, and tens of units) in such a way that the sum is divisible by 3.

- Case 1: Remainder is 0 (divisible by 3)
For a remainder of 0, we need to select three digits whose sum is also divisible by 3. There are two possibilities for this:
a) Selecting three even digits: 0, 2, and 4 (sum = 0+2+4=6)
b) Selecting three odd digits: 1, 3, and 5 (sum = 1+3+5=9)

- Case 2: Remainder is 1
For a remainder of 1, we need to select two even digits and two odd digits. There are two possibilities for this:
a) Selecting two even digits: 0 and 2, and two odd digits: 1 and 3 (sum = 0+2+1+3=6)
b) Selecting two even digits: 0 and 4, and two odd digits: 1 and 5 (sum = 0+4+1+5=10)

- Case 3: Remainder is 2
For a remainder of 2, we need to select two odd digits and two even digits. There is one possibility for this:
a) Selecting two odd digits: 1 and 3, and two even digits: 0 and 4 (sum = 1+3+0+4=8)

Step 4: Calculate the number of ways for each case and sum them up.
a) For case 1, there are 2 options for selecting three even digits and 2 options for selecting three odd digits. Therefore, the total number of ways for case 1 is 2+2=4.
b) For case 2, there are 2 options for selecting two even digits and 2 options for selecting two odd digits. Therefore, the total number of ways for case 2 is 2+2=4.
c) For case 3, there is 1 option for selecting two odd digits and 1 option for selecting two even digits. Therefore, the total number of ways for case 3 is 1+1=2.

Step 5: Sum up the total number of ways for each case.
Total number of ways = 4+4+2 = 10.

Therefore, the correct answer is option 'D' (None of the above), as none of the given options match the calculated total number of ways.
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A five digit number divisible by 3 is to be formed using the numerals 0,1,2,3,4 and 5 without repetition. The total number of ways in which this can be done is:a)220b)600c)240d)None of the aboveCorrect answer is option 'D'. Can you explain this answer?
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