There are 6 different balls and 6 different boxes of the colour same as of the colour of balls then match the column List - I List - II (P) The number of ways in which no ball goes in the box of its own colour is (1) 16 (Q) The number of ways in which atleast 4 balls goes into their own boxes is (2) 40 (R) The number of ways in which atmost 2 balls goes into their own boxes is (3) 265 (S) The number of ways in which exactly 3 balls goes into their own boxes is (4) 664

Engineering

Mathematics

Fundamental Principal of Counting

Question

There are 6 different balls and 6 different boxes of the colour same as of the colour of balls then match the column

List - I	List - II
(P) The number of ways in which no ball goes in the box of its own colour is	(1) 16
(Q) The number of ways in which atleast 4 balls goes into their own boxes is	(2) 40
(R) The number of ways in which atmost 2 balls goes into their own boxes is	(3) 265
(S) The number of ways in which exactly 3 balls goes into their own boxes is	(4) 664

P - 3, Q - 1, R - 2, S - 4

P - 3, Q - 1, R - 4, S - 2

P - 4, Q - 1, R - 3, S - 2

P - 4, Q - 1, R - 2, S - 3

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(P) $6! [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!}] = 265$ [using dearrangement]

(Q) Either four balls go into their correct boxes or all the balls go into the correct boxes

⁶C₄ × 1 + 1 = 16

(R) Either all balls go into incorrect boxes or exactly one ball go into its correct box or two balls go into their

correct boxes

$265 +^{6} C_{1} \times 5! [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}] +^{6} C_{2} \times 9$

265 + 6 × 44 + 135 = 664

(S) ⁶C₃ × 2

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