Consider three boxes, each containing 10 balls labelled \(1,2, \ldots \ldots, 10\). Suppose one ball is randomly drawn from each of the boxes. Denote by \(\mathrm{n}_{\mathrm{i}}\), the label of the ball drawn from the \(i^{\text {th }}\) box, \((i=1,2,3)\). Then, the number of ways in which the balls can be chosen such that \(\mathrm{n}_{1}<\mathrm{n}_{2}<\mathrm{n}_{3}\) is:

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Permutations and Combinations

Consider three boxes, each containing 10 balls labelled

1, 2, \dots\dots, 10

. Suppose one ball is randomly drawn from each of the boxes. Denote by

n_{i}

, the label of the ball drawn from the

i^{th}

box,

(i = 1, 2, 3)

. Then, the number of ways in which the balls can be chosen such that

n_{1} < n_{2} < n_{3}

is:

(A)120✓ Correct
(B)82
(C)240
(D)164

Answer: (A)

Solution

Each box gives a label from

1

10

, and we want strictly increasing labels

n_{1} < n_{2} < n_{3}

. Any choice of three distinct numbers can be arranged in increasing order in exactly one way, so the count of increasing triples is simply the number of ways to choose

3

different numbers from

10

, namely

(3 10) = \frac{10 \cdot 9 \cdot 8}{6} = 120

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Permutations and Combinations

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