Question

Three boys and six girls are being seated in a row of nine chairs on a stage which are numbered from left to right. How many seating arrangements are there if: A. There are no restrictions as to how they are seated? B. The boys sit in the middle three seats? C. Girls sit in the first two seats as well as the last seat? D. The first four seats alternate girl, boy,

Answer 1

Answer:

A. 362,880

B. 4,320

C. 14,400

D. 21,600

Step-by-step explanation:

A.

If there are no restrictions as to how they are seated, then we have permutations of 9 elements and there are 9! (factorial of 9) = 362,880 different seating arrangements.

B.  

If the boys sit in the middle three seats, they can sit in 3!=6 different ways, the girls can sit then in 6!=720 different ways. By the fundamental rule of counting, there are 6*720 = 4,320 different seating arrangements.

C.

We now have arrangements of the type

g, g, x, x, x, x, x, x, g  

The three girls at the ends can be chosen in C(6;3) (combinations of 6 taken 3 at a time) =  

$\large \frac{6!}{3!(6-3)!}=\frac{6!}{3!3!}=20$

different ways. The 6 in the middle can be sit in 6!=720 different ways.

By the fundamental rule of counting, there are 20*720 = 14,400 different seating arrangements.

D.

Now we have arrangements

g,b,g,b,x,x,x,x,x

For the 1st position we have 6 possibilities, for the 2nd we have 3 possibilities, for the 3rd we have 5 possibilities and for the 4th we have 2 possibilities. For the last 5 we have 5!=120 possibilities.

By the fundamental rule of counting, there are 6*3*5*2*120 = 21,600 different seating arrangements.