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Permutations and Combinations question

2025 · 28 Jan · Shift 1 · Q41
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Permutations and Combinations question

2025 · 28 Jan · Shift 1 · Q41

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0 , 1,2,3,4,5,6,71,2,3,4,5,6,71,2,3,4,5,6,7, such that the sum of their first and last digits should not be more than 8 , is
  1. A
    5720
  2. B
    5719
  3. C
    4608
  4. D
    4607
View written solutionFree

Correct answer: 1080

  1. Digits available

    We can form 5-digit numbers using the digits {0,1,2,3,4,5,6,7}\{0,1,2,3,4,5,6,7\}{0,1,2,3,4,5,6,7} Repetition is not mentioned, so we take no repetition.

    Let the 5-digit number be abcdeabcdeabcde where:

    • aaa = first digit
    • eee = last digit
  2. Condition for being greater than 500005000050000

    Since the number must be greater than 500005000050000, the first digit must be one of a∈{5,6,7}.a\in\{5,6,7\}.a∈{5,6,7}.

  3. Condition on first and last digits

    We need a+e≤8.a+e\le 8.a+e≤8.

    Also, since digits do not repeat, e≠ae\ne ae=a.

  4. Count possible (a,e)(a,e)(a,e) pairs

    We check each possible first digit.

    Case 1: a=5a=5a=5

    Then 5+e≤8  ⟹  e≤3.5+e\le 8 \implies e\le 3.5+e≤8⟹e≤3. Possible last digits: e∈{0,1,2,3}e\in\{0,1,2,3\}e∈{0,1,2,3} So, number of choices = 444.

    Case 2: a=6a=6a=6

    Then 6+e≤8  ⟹  e≤2.6+e\le 8 \implies e\le 2.6+e≤8⟹e≤2. Possible last digits: e∈{0,1,2}e\in\{0,1,2\}e∈{0,1,2} So, number of choices = 333.

    Case 3: a=7a=7a=7

    Then 7+e≤8  ⟹  e≤1.7+e\le 8 \implies e\le 1.7+e≤8⟹e≤1. Possible last digits: e∈{0,1}e\in\{0,1\}e∈{0,1} So, number of choices = 222.

    Hence total valid choices for first and last digits: 4+3+2=9.4+3+2=9.4+3+2=9.

  5. Fill the middle three places

    After fixing aaa and eee, we must fill the 2nd, 3rd, and 4th places using the remaining digits.

    From 8 digits, 2 are already used, so 666 digits remain.

    Number of ways to arrange 3 distinct digits in the middle is 6P3=6×5×4=120.{}^6P_3 = 6\times 5\times 4 = 120.6P3​=6×5×4=120.

  6. Total count

    Therefore, total numbers are 9×120=1080.9\times 120 = 1080.9×120=1080.

  7. Check options

    The obtained answer is 1080,1080,1080, which is not among the given options.

  8. Comparison with stored answer

    Stored correct answer is D: 4607, but our correct counting gives 1080.1080.1080.

    So the stored answer appears to be incorrect.

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