JAMB Mathematics 2011 Paper

Concept:This is a permutation problem because the three rooms are distinct and the seven posters are different.Explanation:To satisfy the condition of placing at least one poster in each room, she chooses one poster for the bedroom, one for the living room, and one for the kitchen.For the bedroom, there are $7$ possible posters.For the living room, there are $6$ remaining posters.For the kitchen, there are $5$ remaining posters.Using the multiplication principle:$$7 \times 6 \times 5 = 210$$In permutation form:$$^{7}P_{3} = \frac{7!}{(7-3)!} = \frac{7!}{4!}$$$$= \frac{7 \times 6 \times 5 \times 4!}{4!} = 210$$Therefore, she has $210$ different ways to hang the posters while ensuring each room gets at least one poster.Answer:D. $210$