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Permutations and Combinations question

2023 · 29 Jan · Shift 1 · Q47
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Permutations and Combinations question

2023 · 29 Jan · Shift 1 · Q47

JEE MainMathematicsPermutations and CombinationsNumerical+4 / −1
Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 1436

  1. Understand the ordering

    Five-digit numbers are formed using the digits {1,2,3,5,7}\{1,2,3,5,7\}{1,2,3,5,7} with repetition allowed.

    Since they are arranged in descending order, the largest number gets serial number 111.

    Example: 777777777777777 has serial number 111.

    We need the serial number of 353373533735337.

  2. Idea

    The serial number of 353373533735337 is: 1+(number of 5-digit numbers greater than 35337)1 + \text{(number of 5-digit numbers greater than }35337)1+(number of 5-digit numbers greater than 35337)

    Since each position can be filled with any of 555 digits, we count lexicographically from left to right, with digit priority: 7>5>3>2>17 > 5 > 3 > 2 > 17>5>3>2>1

  3. Count numbers greater than 353373533735337

    We examine digit by digit.

    First digit: 333

    Any number starting with a digit greater than 333 will be larger.

    Greater choices are 7,57,57,5 i.e. 222 choices.

    For each such choice, remaining 444 places can be filled in 545^454 ways.

    2⋅54=2⋅625=12502 \cdot 5^4 = 2 \cdot 625 = 12502⋅54=2⋅625=1250

    Second digit: 555

    Now fix first digit as 333.

    Second digit greater than 555 is only 777, i.e. 111 choice.

    Remaining 333 places can be filled in 535^353 ways.

    1⋅53=1251 \cdot 5^3 = 1251⋅53=125

    Running total: 1250+125=13751250+125=13751250+125=1375

    Third digit: 333

    Fix first two digits as 353535.

    Third digit greater than 333 are 7,57,57,5, i.e. 222 choices.

    Remaining 222 places can be filled in 525^252 ways.

    2⋅52=502 \cdot 5^2 = 502⋅52=50

    Running total: 1375+50=14251375+50=14251375+50=1425

    Fourth digit: 333

    Fix first three digits as 353353353.

    Fourth digit greater than 333 are 7,57,57,5, i.e. 222 choices.

    Remaining 111 place can be filled in 555 ways.

    2⋅5=102 \cdot 5 = 102⋅5=10

    Running total: 1425+10=14351425+10=14351425+10=1435

    Fifth digit: 777

    Fix first four digits as 353335333533.

    Fifth digit greater than 777: none.

    So contribution is 000.

  4. Find serial number

    Hence, the number of numbers greater than 353373533735337 is 143514351435

    Therefore its serial number is 1435+1=14361435+1 = 14361435+1=1436

  5. Final answer

    1436\boxed{1436}1436​

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