Answer to Question #140396 in Algebra for Cameron Foster

Question

1. Five girls and four boys are randomly seated in a row. How many seating arrangements are possible if . . .
(a) there are no restrictions?
(b) boys and girls alternate?
(c) boys sit together, girls sit together?
(d) Compa must sit in the middle chair?
(e) Girls sit at the end seats?


2. Please determine whether a permutation, combination, or neither is used. Please set up and solve each problem.
(a) In how many ways can a six card hand be drawn from a standard deck of fifty-two cards?
(b) How many codes are comprised of two letters followed by three digits?
(c) A choral group is known for fifteen songs. How many musical arrangements are possible if the
group has time for five songs?
(d) Compa’s wardrobe contains six sweaters, eight skirts, four headbands, and three pairs of shoes.
How many different outfits can Compa make from her wardrobe?
(e) How many committees of four be formed from fifteen people?

Answer 1

1

(a) In this case, we simply use the permutation formula.

P₉=9!=362880

(b) If boys and girls alternate, then we can seat boys in P₄ ways and girls in P₅ ways. Since these events are simultaneous we get:

P₄*P₅=4!*5!=2880

(c) If the boys sit together we can get it P₄ ways and the girls P₅. Since we can swap them, we get:

2*P₄*P₅=2*4!*5!=5760

(d) Since the Compa sits in the middle, the rest of the guys can be seated in P₈ ways.

P₈=8!=40320

(e) We can seat the girls in the extreme seats A²₅ (since order is important to us, we use the formula of mixing A^mn=n!/(n-m)!). The rest of the guys are P₇ ways. Then we get

A²₅*P₇=(5!/3!)*7!=100800

2

(a) In this example, we use the combination formula C^m_n=n!/(n!*(n-m)!), because we do not care about the order of the combination, since in the end we get a certain combination.

C⁶₅₂=52!/(6!*(52-6)!)=20358520

(b) We use the arrangement formula A^m_n=n!/*(n-m)!, because in this example, the order is important. It turns out that the code from the letters is found in A²₂₆ ways, and the code from the numbers in A³₁₀ ways. Therefore

A²₂₆*A³₁₀=(26!/(26-2)!)*(10!/(10-3)!)=650*720=468000

(c) We use the arrangement formula Amn=n!/*(n-m)!, because in this example, the order is important. We get that we can make arrangements of A⁵₁₅ using the following methods.

A⁵₁₅=15!/10!=360360

(d) In this example, we can take a sweater in 6 ways, a skirt-8, a headband-4, a pair of shoes-3. Thus we get a combination of outfits equal to 6*8*4*3=576 ways.

(e) In this example, the order is not important and there are no repetitions, so we use the combination formula C^m_n=n!/(n!*(n-m)!).

C⁴₁₅=15!/(4!*(15-4)!)=1365