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Permutations and Combinations question

2019 · 12 Jan · Shift 1 · Q32
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Permutations and Combinations question

2019 · 12 Jan · Shift 1 · Q32

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :
  1. A
    164
  2. B
    240
  3. C
    82
  4. D
    120
View written solutionFree

Correct answer: D

  1. Interpret the experiment

Each box contains balls labelled 1,2,…,101,2,\dots,101,2,…,10.

One ball is drawn from each box, so a choice is an ordered triple:

(n1,n2,n3)(n_1,n_2,n_3)(n1​,n2​,n3​)

where each ni∈{1,2,…,10}n_i \in \{1,2,\dots,10\}ni​∈{1,2,…,10}.

We need the number of triples satisfying

n1<n2<n3.n_1<n_2<n_3.n1​<n2​<n3​.
  1. Convert the condition into a selection problem

If

n1<n2<n3,n_1<n_2<n_3,n1​<n2​<n3​,

then the three labels must be distinct.

So we simply need to choose any 3 distinct numbers from {1,2,…,10}\{1,2,\dots,10\}{1,2,…,10}, and once chosen, their increasing order automatically determines:

  • the smallest as n1n_1n1​
  • the middle as n2n_2n2​
  • the largest as n3n_3n3​

Thus, the number of valid ways is

(103).\binom{10}{3}.(310​).
  1. Compute
(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.
  1. Check options
  • A: 164164164 ❌
  • B: 240240240 ❌
  • C: 828282 ❌
  • D: 120120120 ✅

So the correct option is

120.\boxed{120}.120​.
  1. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

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