Solve 3 Digit Odd Number Question with 1, 2, 3, 4, 5, 6, 7

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In summary, for part 1, the total number of 3-digit odd numbers that can be obtained from the digits 1, 2, 3, 4, 5, 6, 7 without repetition is 120, and for part 2, the total number of 3-digit odd numbers that can be obtained from the digits 1, 2, 3, 4, 5, 6, 7 with repetition allowed is 196.
  • #1
jinx007
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Badly need help to attempt this question because i am 90% sure a question like this will appear in the exams that i will be taking next week.

Find out how many 3 digit odd number that can be obtained form digits 1, 2, 3, 4, 5, 6, 7 if

1/ repetition of digit not allowed

2/ Repetition of digit allowed

ANSWER: 1/ 120 2/ 196


MY ATTEMPT

1/

- - - I'VE CONTROLLED THE LAST BLOCK 1, 3, 5, 7

6p2 X 4 = 120

2/

--- I'VE CONTROLLED THE LAST BLOCK 1, 3, 5, 7

AS DIGIT REPEAT

7P2 X 4 = 148

i AM A BIT STUCK IN THE SECOND PART BUT PLEASE VERUFGY THE FIRST PAR AND HELP ME TO ATTEMPT THE SECOND PART
 
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  • #2
. For part 1, your answer is correct. For part 2, the correct answer is 196. The idea is that you consider each digit in the three-digit number separately, and then multiply the results together. For the last digit, you have 4 choices (1, 3, 5, 7). For the middle digit, you have 7 choices, as repetition of digits is allowed here.For the first digit, you have 7 choices, as repetition of digits is also allowed here. Therefore, the total number of 3-digit odd numbers that can be obtained with repetitions allowed is 4 x 7 x 7 = 196.
 

FAQ: Solve 3 Digit Odd Number Question with 1, 2, 3, 4, 5, 6, 7

1. What is the question asking me to solve?

The question is asking you to find a three-digit odd number using the given numbers 1, 2, 3, 4, 5, 6, 7.

2. Are there any restrictions on how the numbers can be used?

Yes, the numbers must be used in order and cannot be repeated. So the first digit must be 1, the second digit must be 2, and so on.

3. How many possible solutions are there?

There are 120 possible solutions. This can be calculated by using the formula n!/(n-r)! where n is the total number of options (7 in this case) and r is the number of options to choose (3 in this case). So, 7!/(7-3)! = 7!/4! = 7*6*5 = 210. However, since we are looking for odd numbers, we need to subtract the numbers that end in 0, 2, 4, or 6, leaving us with 120 possible solutions.

4. Can I use any mathematical operations to find the solution?

Yes, you can use addition, subtraction, multiplication, division, and exponentiation to combine the given numbers and create a three-digit odd number. However, you cannot use any other numbers or mathematical operations.

5. Is there a specific method or strategy to solve this question?

There are several strategies you can use to solve this question. One method is to start with the largest possible number (765), and then work your way down by subtracting 2 from the last digit until you find an odd number. Another method is to focus on the last two digits, which must add up to an odd number, and then use trial and error to find the first digit that creates a three-digit odd number.

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