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Permutations and Combinations question

2025 · 2 Apr · Shift 1 · Q42
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Permutations and Combinations question

2025 · 2 Apr · Shift 1 · Q42

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :
  1. A
    360
  2. B
    2520
  3. C
    1820
  4. D
    45
View written solutionFree

Correct answer: B

  1. We need sequences of length 101010 formed using only 0,1,20,1,20,1,2 such that:

    • exactly five terms are 111
    • exactly three terms are 222

  2. Since there are 101010 total positions, the remaining positions must be 000.

    Number of zeros =10−5−3=2=10-5-3=2=10−5−3=2.

  3. So we must arrange the multiset: {1,1,1,1,1,2,2,2,0,0}\{1,1,1,1,1,2,2,2,0,0\}{1,1,1,1,1,2,2,2,0,0}

  4. The number of distinct sequences is the number of permutations of these 101010 symbols with repetitions: 10!5! 3! 2!\frac{10!}{5!\,3!\,2!}5!3!2!10!​

  5. Compute: 10!5!3!2!=3628800120⋅6⋅2=36288001440=2520\frac{10!}{5!3!2!} = \frac{3628800}{120\cdot 6\cdot 2} = \frac{3628800}{1440} = 25205!3!2!10!​=120⋅6⋅23628800​=14403628800​=2520

  6. Hence the required number of sequences is 252025202520

  7. Checking options:

    • A: 360360360
    • B: 252025202520 ✔
    • C: 182018201820
    • D: 454545

Therefore, the correct option is B.

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