Question

finite mathematics-

A Group of six friends consisting of 4 females and 2 males are seated in a row. How many different seating arrangements are possible

a) If There are no restrictions of seating order?

b) If all for a female sit together?

c) If the order is MWWMWW?

Answer 1

Solution :-  

(A) Total Friends = 6

If there are no restriction  

then Simple Seating Arrangements = 6! = 6 * 5 * 4 * 3 * 2 * 1 = 720 Arrangements

(B)  If all for a female sit together = then the arrangement for females = 4! = 4 * 3 * 2 = 24

Now Treat all Womens As one   then total arrangement done for 3 :- One women group and two single man

= 3! = 3 * 2 = 6

Now total arrangements = 24 * 6 = 144

(C) Now when Group is that no Men together means =  MWWMWW

Arrangement of four women = 4! = 4 * 3 * 2 = 24

Now Arrangement for two men in the middle = 2 ( As only change in both position )

So total Arrangements = 24 * 2 = 48

Now total arrangements = 48

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