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Permutations and Combinations question

2002 · Shift 0 · Q94
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Permutations and Combinations question

2002 · Shift 0 · Q94

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Total number of four digit odd numbers that can be formed using 0, 1, 2, 3, 5, 7 (using repetition allowed) are :
  1. A
    216
  2. B
    375
  3. C
    400
  4. D
    720
View written solutionFree

Correct answer: D

  1. We need to form 4-digit odd numbers using the digits {0,1,2,3,5,7}\{0,1,2,3,5,7\}{0,1,2,3,5,7} with repetition allowed.

  2. Conditions for a 4-digit odd number:

  • The first digit cannot be 000.
  • The last digit must be odd.
  1. Count choices position-wise.

First digit

Possible digits are 1,2,3,5,71,2,3,5,71,2,3,5,7. So, number of choices = 555

Last digit

To make the number odd, the last digit must be one of 1,3,5,71,3,5,71,3,5,7 So, number of choices = 444

Middle two digits

For each of the 2 middle places, any of the 6 digits can be used since repetition is allowed: 6 choices each6 \text{ choices each}6 choices each

So total choices for the middle two digits: 6×6=366 \times 6 = 366×6=36

  1. Total number of such numbers: 5×6×6×4=5×36×4=7205 \times 6 \times 6 \times 4 = 5 \times 36 \times 4 = 7205×6×6×4=5×36×4=720

  2. Therefore, the total number of four-digit odd numbers is 720\boxed{720}720​

  3. Comparing with the stored correct answer:

  • Stored correct answer: D
  • Derived answer: D

Hence, the derived answer agrees with the stored answer.

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