talaangoshtari wrote:Hi GMATGuruNY,
Would you please check my answer?
Eight boys have to be seated in eight chairs numbered 1 to 8 in a row. In how many ways this seating can be done such that if a particular boy does not want to sit in the even numbered chair and another boy wants to sit at the extremes?
4×6! + 4×6! = 8×6!
Let E = the boy who must sit at either end and O = the boy who must in an odd-numbered chair.
Case 1: E in seat 1
Number of options for O = 3. (Seat 3, 5 or 7.)
Number of ways to arrange the remaining 6 boys = 6!.
To combine these options, we multiply:
(3)(6!).
Case 2: E in seat 8
Number of options for O = 4. (Seat 1, 3, 5 or 7.)
Number of ways to arrange the remaining 6 boys = 6!.
To combine these options, we multiply:
(4)(6!).
Total ways = (3)(6!) + (4)(6!) = [spoiler](7)(6!)[/spoiler].
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