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Permutations and Combinations question

2006 · Shift 0 · Q72
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Permutations and Combinations question

2006 · Shift 0 · Q72

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
At an election, a voter may vote for any number of candidates, not greater than the number to be elected. There are 10 candidates and 4 are of be selected, if a voter votes for at least one candidate, then the number of ways in which he can vote is
  1. A
    5040
  2. B
    6210
  3. C
    385
  4. D
    1110
View written solutionFree

Correct answer: C

  1. Understand the voting rule

A voter may vote for any number of candidates, provided that number is not greater than the number to be elected.

  • Total candidates = 101010
  • Number to be elected = 444
  • Voter must vote for at least one candidate

So the voter can choose:

  • 111 candidate, or
  • 222 candidates, or
  • 333 candidates, or
  • 444 candidates
  1. Count the number of ways

If the voter votes for exactly rrr candidates, the number of ways is: (10r)\binom{10}{r}(r10​)

Hence total number of ways is: (101)+(102)+(103)+(104)\binom{10}{1}+\binom{10}{2}+\binom{10}{3}+\binom{10}{4}(110​)+(210​)+(310​)+(410​)

  1. Compute each term

(101)=10\binom{10}{1}=10(110​)=10

(102)=10⋅92=45\binom{10}{2}=\frac{10\cdot 9}{2}=45(210​)=210⋅9​=45

(103)=10⋅9⋅83⋅2⋅1=120\binom{10}{3}=\frac{10\cdot 9\cdot 8}{3\cdot 2\cdot 1}=120(310​)=3⋅2⋅110⋅9⋅8​=120

(104)=10⋅9⋅8⋅74⋅3⋅2⋅1=210\binom{10}{4}=\frac{10\cdot 9\cdot 8\cdot 7}{4\cdot 3\cdot 2\cdot 1}=210(410​)=4⋅3⋅2⋅110⋅9⋅8⋅7​=210

  1. Add them

10+45+120+210=38510+45+120+210=38510+45+120+210=385

  1. Match with the options

385385385 corresponds to Option C.

  1. Compare with stored correct answer

Stored correct answer = C

Our derived answer = C

So the stored answer is correct.

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