Question

Suppose 7 men and 6 women attended a concert and their tickets were together in one row. How many ways could they have been organized so that: a. they were randomly assigned seats? b. the men and women alternated so that men sat by women and women sat by men? c. all men sat together, and all women sat together?

Answer 1

Answer:A)13!=6227020800 ways

B)6!×7!=3,628,800ways

C)If the men sit together,=10,080 ways

If the women sits together=1,440ways

Explanation:to arrange 7 men and 6 women randomly in a straight line using permutation is 13p13= 13factorial=13!=13×12×11×10×9×8×7×6×5×4×3×2×1=622702080 ways of arranging them randomly i.e in no specific order.

B) Arranging 7 men and 6Women such that the women and men alternate,let's first assign the women this will require 6! Ways of arranging men,then the first man will have 7ways,next 6 choices,next man will have 5 choices,and so on.so number of ways of arranging men in alternate to women=7 ×6×5×4×3×2×1=5040

So total ways of alternate arrangements=5040×6!=1260×6×5×4×3×2×1=3,628,800ways

C)if men will seat together there will be 7!ways while the women will all seat to their left or to their right(2ways).

So there 7!×2ways=5040×2=10080ways

II) If the women will seat together ,it will be 6!×2=6×5×4×3×2×1×2=1440ways

Answer 2

(a) there are 13! factorial ways that they could have been assigned seats;

that is 13x12x11x10x9x8x7x6x5x4x3x2z1=6 227 020 800 ways

(b) the tendency of a man sitting next to a woman and a woman sitting next to a man = 7!=7x6x5x4x3x2x1=5 040 ways

(c) all men + 7!= 7x6x7x5x4x3x2x1= 5 040 ways

also all women sat together= 6x5x4x3x2x1=720 ways.

Step-by-step explanation:

for the siting arrangement we use combination ; where the sitting arrangement does not matter because there is no preference.