Problem 13, February 1995Author(s): Wayne PulferSource: The Mathematics Teacher, Vol. 89, No. 6 (SEPTEMBER 1996), pp. 502-503Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27969859 .

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-2(3)2+?(3) -18 + 18 (3)+ 3

= 0.

QED.

Having reached an identity, we have proved that for the limit as

goes to 3, ( 2 - 9 3 = 1. Using

the same technique, it can also be

proved that for the general problem,

Solve for x:

where f(x) andgte) are polynomi als of finite degree, any solution to the equation g(x) = 0 is a solu tion to the problem regardless of whether the base is also equal to zero. So despite my father's objec tions and my initial concern, the solution was exactly correct in this assertion.

The question that naturally arises is "Are any examples of zero raised to the power of zero not equal to 1?" The answer is yes, but the base must go to zero faster than any power of x. For

example,

lim x->0|

f ! Y

\ellxj

=.A.

Taking the natural log of both sides and pulling out the limit give

lim \ In x->0

' *

\eVxJ

= ln(A).

Doing algebra leads to

limixlnl x->0

= ln(A),

limj-*ln(e1/:?)} =

ln(A),

lim{-l} =

ln(A),

-l = ln(A),

A = e-1 = 0.368.

Edward Forringer Forringe@pilot. msu. edu 2310 Prince Street Durham, NC 27707

Foster responds: The objec tions submitted by Forringer regarding question 8 in the Jan uary 1996 issue are certainly valid. However, we must keep in mind that the intent of many of these problems is to keep them

simple enough so that many stu dents are able to solve them with minimal assistance from their mathematics teacher.

Problem 13, February 1995

Problem 13, January 1996 Problem 13 in the February 1995 Calendar asks, "In how many ways can a committee of four

people be chosen from five mar ried couples if no committee is to include a husband-and-wife

pair?" Problem 13 in the January 1996 Calendar asks, "Given twenty couples, how many differ ent three-member committees can be formed that do not contain any couple?"

The solution given for the Feb ruary problem is (5C4)24 = 80. This result can be interpreted as

choosing four couples and then choosing one person from each couple. As I pointed out in "Read er Reflections" on page 509 of the September 1995 issue, this answer can also be written (10C4)

- 5(8C2) +

(5C2) = 80. The solution for the January problem is (40C3)

-

38(200^ = 9120. As can be seen from the February problem, this answer can also be written

(20C3)23 = 9120. Yet another way is available to

approach these problems. For the

February problem, ten choices exist for the first person on the committee, eight for the second, six for the third, and four for the fourth. Therefore, the number of committees is (10 ? 8 ? 6 ? 4)/4L The division by 4! is needed be cause the order of the committee members does not matter. Simi

larly, for the January problem, forty choices exist for the first member, thirty-eight for the sec

ond, and thirty-six for the third; the number of committees is (40 ? 38 ? 36)/3! = 9120. Extend ing the January problem to a committee of four gives

(40 ? 38 ? 36 ? 34)/4! = 77 520, which equals the answer that can be gotten by other methods and suggests a recurrence relation.

Let Un be the number of 72-member committees to be formed from twenty couples, with no committee containing a cou

ple. Then U1 = 40, and for > 1, = ( ̂ (40-2.( -1)))/ . For

the February problem, let Vn be the number of rc-member commit tees to be formed from five couples with no committee containing a

couple. Then Vl = 10 and Vn =

VnA (10 -2(n- l))ln. These equations can easily be entered into a TI-82 calculator in Seq

mode to create a table of values for Un and Vn. It is interesting to note that these recurrence rela tions work for all positive n. For

> 20, Un = 0; for > 5, Vn = 0. Clearly, for m couples, the recur sion equation becomes = 2m imdUn = (Un_1(2m-2(n-l)))/n.

George M. Caplan george. caplan@channell. com

One Colonial Terrace Belmont, MA 02178-2970

Problem 13, February 1995

In how many ways can a com

mittee of four people be chosen from five married couples if no committee is to include a husband-and-wife pair?

The solution given was (5C4)24 = 80. George Caplan, in "Reader Reflections" (September 1995, 510), provided a second solution: (10C4)-5(8C2) + (5C2) = 80. How ever, for this problem I chose to give my Finite Mathematics stu dents these two solutions plus two others. The problem then became one of explaining the rea soning behind each solution. This approach encourages students to do more than just give their "numerical" answers.

II. iqC4 - 5? 8 C 2 ) +5

C 2 was given as

L(5C4)2 was given as

III.

5 ?Hi

IV. 10x8x6x4

4!

I prefer the bracket notation

instead of nCr or G(n, r). Having to bracket 5(8C2) does not seem as "clean" as

or

Solution III introduces cases and symmetry. The expression

'2 1

could be five men choose three, two nonwives choose one or five women choose three, two non

husbands choose one. Solution IV suggests that a

"combination" problem could be done as a "permutation" problem.

I offer a final thought. As has often been suggested in your pub lication, having students write about mathematical ideas and results is a crucial part of the learning experience. Perhaps

writing is especially important when dealing with ?Cr and rtPr, that is, combinations and permu tations, for example,

Explain why

71-1 n-1

r-1

A possible answer follows:

Left side: number of r-subsets from a set of objects.

Right side:

~ number of r-subsets with one

particular element excluded.

(n-1) [r-1)

~ number of r-subsets with the same particular element included.

n-1 n-1

r-1

502 THE MATHEMATICS TEACHER

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must equal ( \

Two levels of understanding really do exist. What is C(8,3)? The answer is the number of 3-subsets from 8. What is the value of C(8,3)? The answer is

iL 5!3!*

Wayne Pulfer Adult High School Ottawa, ON KIR 7N4

Problem 26, January 1996 My first-year-algebra class devel oped another solution to the problem without using the con cept of the center of the square. ABCD is a unit square with D at(l, 0) and Cat (2, 0). Find the equation of the line through the origin that bisects the area of this square.

Using the solution's diagram,

A

we know that the area of square ABCD is 1; hence the area of trapezoid EFCD is 1/2. We shall consider the height of the trape zoid to be DC, or 1, and the bases to be DE and CF. So

Area of trapezoid EFCD

= ^ ? (sum of the

bases) ? height

2 2 1 '

We conclude that DE + CF= 1. Let DE = j and CF = k. The coor

dinates of E are therefore (1 J), whereas the coordinates of F are (2, k). Because the origin, E, and

F are collinear, the slope of the line between any two of them is constant. So

j-Q=k-0=k-j l-0~2-0~2-l

or

j = k_k-j 12 1

Working with the first propor tion, we know that j = 2k. Remembering that j + k = 1, we solve the system of equations

\k = 2j l/ + * = l

to determine that j - 1/3 and k - 2/3. The coordinates of E are, therefore, (1,1/3), and the deter mination of the equation of the line OE follows directly.

David M. Early dearly@lfa. Ifc. edu Lake Forest Academy Lake Forest, IL 60045

Problem 9, February 1996 Here is an alternative solution to the 9 February 1996 Calendar problem.

Find the value of s:

It encourages students to use their intuitive problem-solving strengths rather than to search for "tricks":

V 7

1 1 1 + - + ? + ? +

9 27 81

( 1 V 6 7

+ V is;

1

+

1 1

= 3

4

Angelo DeMattia Columbia High School

Mapkwood, NJ 07040

Problem 12, March 1996 I I found two problems with your solution to the 12 March Calen dar problem.

Larry's birthday is in April. Michelle's birthday is in Octo ber. Determine the probability that their birthdays are within 180 days of each other.

First, if I said that my birthday was within one day of yours, and mine was 24 May, then yours could be 23,24, or 25 May. Thus, the open sentence should be written as y - < 180, not just less than.

Second, just as the sine func tion repeats every 2 , the calendar repeats each 365 days, excluding leap year. So Michelle's birthday may follow Larry's birthday that calendar year and be within 180 days, or

Larry's birthday may be in the next calendar year, following Michelle's within 180 days. The integer is in the interval [91,120] or [456,485], and y is in the interval [274,304]. In the first interval for x, day 94, 4 April, is the first day that will pair with day 274; 1 October, day 95, pairs with days 274 and 275, and so on, to satisfy \y-x \ < 180. This course gives 1 + 2 + 3 + ... + 26 + 27 = 378 possible two day pairs when October follows April. In the second interval for

, day 276,3 October, pairs with day 456,1 April of the year fol lowing; day 277 pairs with days 456 and 457, and so on, satisfying |y-x| <180. This plan of action yields 1 + 2 +

3 + . ?. + 28 + 29 = 435. The total "hits" versus total possibilities is (378 + 435V930 = 813/930 = 271/310 - 0.874. Therefore, the probability that their birthdays are within 180 days of each other is about 7/8.

This problem could also have been solved by determining the pairings of dates where \y-x \ > 180 and then subtracting that result from 1.

The article by Sen, Raiszadeh, and Raiszadeh contains a typo

graphical error in the last equa tion on page 198, column 1. It should read

not

The calculations were done correctly.

Brian H. Sjolseth P. 0. Box 163342 Austin, TX 78716-3342

Problem 12, March 1996 II Please allow me to comment on the 12 March Calendar problem and proposed solution in the March 1996 Mathematics Teacher. If an event is to occur within two days of, say, Wednesday, 17 July, then the event must occur on one of five days: 15,16,17,18, or 19 July. If the event occurs on 14 July or 20 July, then it did not occur within two days of 17 July. Thus, two or more full days can not occur between the pivotal date, 17 July, and the actual date of occurrence, which could be on either side of the pivotal date.

Between Larry's birthday in April and Michelle's birthday in October is a "summer" span and a "winter" span.

Summer Span May 31 June 30

July 31 August 31

September 30

Total 153

Winter Span November 30 December 31

January 31

February 28 March 31

Total 151

The two birthdays could occur on 30 31 = 930 possible pairs of (April, October) dates. I suggest counting the impossible pairs and then subtracting from 930 to get the possible pairs for which we can say that their birthdays are "within 180 days of each other."

If we rule out leap year, then 365 - 2 (for the actual birthdays)

Vol. 89, No. 6 ? September 1996 503

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