Permutations and combination

Question

Speech of 7 politicians in a political rally is arranged one after one other, number of ways in which they can do the same if-

(i) No restriction

(ii) Mr. A wants to speech before Mr. B and Mr. B before Mr. C.

(iii) Mr. A wants to speech before Mr. B and Mr. B after Mr. C.

Answer 1

(i) If there are no restrictions, the number of ways to arrange the speeches of 7 politicians is simply the number of permutations of 7 distinct objects. This is given by 7!.

7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 = 5040

(ii) For Mr. A before Mr. B and Mr. B before Mr. C,

A must come before B, and B must come before C. The remaining 4 politicians can be arranged freely.

• Without restrictions, there are 7! ways.
• Among the 3 politicians A, B, C any of the 3! = 6 arrangements are possible.
• However, only 1 of these 6 arrangements satisfies A before B and B before C.

\(\frac{7!}{3!}\) = \(\frac{5040}{6}\) ​= 840.

(iii) For Mr. A before Mr. B and Mr. B after Mr. C,

\(\frac{7!}{3!}\) = \(\frac{5040}{6}\) = 840