Respuesta :
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) =
[tex]\large \frac{6!}{3!(6-3)!}=\frac{6!}{3!3!}=20[/tex]
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.
To answer this question, it is necessary to use Combinations Theory, more precisely the concept of Permutations.
The solution is:
A) P₉ = 362880
B) Pr = 432
C) Pc = 64800
D) Pf = 4320
A) To compute the number of arrangements of 6 boys and 3 girls, we just calculate the permutations of 9 elements ( without any restriction) then
Permutations of 9 elements:
Pn = n!
P₉ = 9!
P₉ = 9×8×7×6×5×4×3×2×1
P₉ = 362880
B) Restriction: The three boys will be seated, in the middle three seats (Pr)
In this case, we have two groups of elements:
- Three boys P₁
- Six girls P₂
To compute the way in which three boys could be seated
P₁ = 3!
P₁ = 3×2×1
P₁ = 6
The group of girls, must be treated as follows
The girls need to be split into two groups, ( only in that case the boys will occupy the three middle seats) let´s say group A and group B
We located group A at the beginning of the row, we get
Pₐ = 3! Pₐ = 6
and the group at the end
P₂ = 3! P₂ = 6
For each one of the combinations, of group A, we will get 6 times of different combinations then Pₐ×P₂
Pₐ×P₂ = 36
And as we need to considerer changing the positions of the groups, we multiply Pₐ×P₂ ×2
Pₐ×P₂ ×2 = 72
Finally, Pₐ×P₂ ×2 has to be multiplied by P₁, giving us the answer for this case
Pr = 72×6
Pr = 432
C) Restriction: Girls will be seated 2 at the beginning and 4 at the end
We calculate the number of combinations for the boys P = 3!
P = 6
And for the first group, (the beginning) of girls, variations of 6 elements taking 2 at the time
V (m,n) = m! / (m-n)! = 6! / ( 6 - 2 )! = 6!/4!
V (6,2) = 6×5
V(6,2) = 30
For the last group, V(6,4) = 6!/ (6 -4)!
V(6,4) = 6!/ 2!
V(6,4) = 360
Finally the number for that restriction is:
Pc = 6×30×360
Pc = 64800
D) Restriction: The first four seats alternate girls and boys
we have 6 ways of alternate boys and girls
b₁ g₁ b₂ g₂ b₂ g₁ b₁ g₂ b₁ g₂ b₂ g₁
g₁ b₁ g₂ b₂ g₂ b₁ g₁ b₂ g₁ b₂ g₂ b₁
And the rest, 4 girls and 2 boys will permute freely then
P = 6!
P = 6×5×4×3×2×1
P = 720
To get the number we multiply
720 × 6
Pf = 4320
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