First slide

Combinations

Question

$In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in$
$such a way that each of these rooms contains at least one person and at most two persons.$
$Then the number of all possible ways in which this can be done is _____$

Difficult

Solution

$4 Rooms \to 6 People 6 \to \underline{1} \underline{1} \underline{2} \underline{2}$
$\frac{6!}{(1)! 1! 2! 2! 2! 2!} \times 4! = 1080$

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