Possible Solutions: Test #2

1a) 14€13€12€11

1b) 14!/10!

1c) C(14,4)€4!

2) 9!/(1!4!3!1!)

There are 9! ways to arrange the 9 letters in the word. We
divide by
(1!4!3!1!) to account for the arrangements of identical
letters that
result in no visible rearrangement of the letters.

3a) C(20,3)

3b) 1140

4) C(18,9)=C(17,9)+C(17,8)

Pascal's Formula states that C(n+1,k+1)=C(n,k+1)+C(n,k).

5) 46 jokes

If Juanita works every night for the next 40 years, there are
at
least 14610 nights, allowing for leap year every four years
(40€365+10).
We then need to determine a value for n so that C(n,3)„14610.
By
trial and error or by solving the inequality
n!/[3!(n-3)!]>=14610, over the positive integers, we
determine that n„46 satisfies the desired relationship.

6) Case I of the proposed solution is accurate. For Case II, the solution as presented uses an incorrect assumption that the left-most digit will always be the digit that is greater than 7. This is more restrictive than stated in the original problem.

To rectify this, we should consider the two other possible positions for the digit greater than 7. In each case, the argument as stated will show there are 2€8€7 possible arrangements. So for Case II, there are three times as many possibilities as suggested by the solver.

Therefore, there are 8€7€6 + 3(2€8€7) possible arrangements for the briefcase lock.

7a) 6

In the expansion, the variable a will appear as a factor from
0 to 5
times in any one term.

7b) 32*a*^{5} + 80*a*^{4}*b* +
80*a*^{3}*b*^{2}

This is simplified from C(5,0)(2a)^{5}(b)^{0}
+ C(5,1)(2a)^{4}(b)^{1} +
C(5,2)(2a)^{3}(b)^{2}.

8a) 4^{25}

For each of 25 factors (e + f + g + h), there are four terms
to
choose from to carry out the expansion.

8b) C(28,3)

Each collected term must be of the form
[25!/(E!F!G!H!)]e^E€f^F€g^G€h^H,
where E+F+G+H=25 solved over the nonnegative integers. There
are
C(25+4-1,4-1) solutions to this equation.

8c) 25!/(1!6!10!8!)

Refer to the general representation for a collected term,
shown in
(8b).

9) [C(6,4)€C(16,12)]€[C(7,4)€C(15,12)]

George must pass through the bakery. There are C(6,4)
different
6-block routes he can take. He then must go 16 blocks to work,
including 12 westerly blocks. There are C(16,12) routes from
the
bakery to work. Using the same reasoning for the return trip,
there
are C(7,4) routes to the bank from work and C(15,12) routes
from the
bank to George's home. Because each route to work can be
matched with
each route from work, we multiply to get the total number of
round-trip paths George may take.

10) C(12,5)

Line up the red books first. There are 12 spaces among the red
books,
into which we must place the 5 green books. This can be done
in
C(12,5) ways. We use combinations rather than permutations
because we
do not concern ourselves with either how the 11 red books nor
the 5
green books are arranged among themselves. The books are
distinguishable only by color.