Permutation and CombinationHard
Question
There are 10 seats in the first row of a theatre of which 4 are to be occupied. The number of ways of arranging 4 persons so that no two persons sit side by side is:
Options
A.7C4
B.4 . 7P3
C.7P3. 4!
D.840
Solution
x1 + x2 + x3 + x4 + x5 = 6
x1 + y1 + y2 + y3 + x5 = 3
but x1, x5 ≥ 0
x2, x3, x4 ≥ 1
⇒ y1, y2, y3 ≥ 0
3 + 5 - 1C3 . 4 ! = 7C3 . 4!
= 7p3 . 4 = 840
x1 + y1 + y2 + y3 + x5 = 3
but x1, x5 ≥ 0
x2, x3, x4 ≥ 1
⇒ y1, y2, y3 ≥ 0
3 + 5 - 1C3 . 4 ! = 7C3 . 4!
= 7p3 . 4 = 840
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