A probability problem by Aman Goenka

This solution is for all beginners, completely explained reasoning.

As we want total nights to be $6$ , the number of nights for which friend was invited can be $(3,3)$ or $(3,2,1)$ or $(2,2,2)$

These are the only cases.

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Number of ways in which each case is possible is $w_f \times w_n \times w_a$ where $w_f =\text{ways of choosing the friends} \\ w_n = \text{ways of assigning them the number of nights} \\ w_a = \text{ ways of arranging their nights among the 6}$

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$\mathbf{1.}$ $(3,3)$

Ways of choosing 2 friends for inviting- $\binom{3}{2}=3$

Ways of assigning them the number of nights- 1 (giving 3 to each)

Ways of arranging them in those 6 nights - $\binom{6}{3} \times \binom{3}{3}=20$

Ways in which the case is possible - $3\times 1\times 20=60$

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$\mathbf{2.}$ $(3,2,1)$

Ways of choosing friends for inviting = $\binom{3}{3}=1$

Ways of assigning them the number of nights (3 or 2 or 1) = $3!=6$

Ways of arranging their invites over the 6 nights - $\binom{6}{3} \times \binom{3}{2}$

Ways in which the case is possible- $6\times 20\times 3 = 360$

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$\mathbf{3.}$ $(2,2,2)$

Ways of choosing friends - $\binom{3}{3}=1$

Ways of assigning them the number of nights - $1$ (giving 2 to each of them)

Ways of arranging their invites over the 6 nights - $\binom{6}{2} \times \binom{4}{2} =15\times 6=90$

Ways in which the case is possible $1\times 1\times 90 = 90$

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Hence total number of ways $60+360+90=\boxed{510}$