1. While lounging in the lobby of the Embassy Suites hotel in San Diego, I watched hotel employees open a door after entering a code by pushing a digital keypad. The keypad is similar to the one shown below. From my vantage point, I could see neither the door nor the keypad. I could see people pushing buttons from the side. After carefully watching several people enter the door, I concluded that the sequence of keystrokes they used looked like the figure shown below. I knew the order of the keystrokes, as indicated by "first, second" and so on shown here. I did not know which keypad was struck first. I did know that the fourth push was directly below the first and second, and that the third and fifth pushes were directly below the fourth.
Based on this information, determine the number of different 5number codes that could possibly unlock this door.

There are three possible keypads that could have been pushed first. This generates 6 different possible first numbers. The second through fifth pushes each have 2 possible numbers associated with them. This results in 2^4 = 16 different 4number sequences. Together, then, there are 6*16 = 96 possible 5digit codes.
Consider worstcase scenarios, trying to place pennies so that the condition is not met.
(i) To keep pennies in less than five columns, I can use no more than 40 pennies (4 columns by 10 rows). As soon as the 41st penny is placed, by the pigeonhole principle we are assured that at least 5 columns and 5 rows are occupied, thereby providing a placement from which we can choose a set of 5 pennies that meets the condition described above.
(ii) If any empty grid squares exist in a set of 4 columns, a penny must exist in a fifth column. Given this condition, 5 different rows must have pennies in them, for if not, the pennies would only be in 4 different rows, which by (i) above is impossible.
(iii) By symmetry, the words "rows" and "columns" can be interchanged in (i) and (ii) with no loss of generality.
See solution to question from Spring 99 sample exam questions!
Your task: Justify that this relationship will hold in general, or, alternatively, show that the result cannot be generalized to all of Pascal's triangle.
By brute force symbolic manipulation we can show that
C(n,k2) + C(n,k1) + C(n,k)  [ C(n3,k3) + C(n2,k3) + C(n1,k3) ]  [ C(n3,k2) + C(n2,k1) + C(n1,k1) ] = C(n2,k2) + C(n1,k2) + C(n1,k1).
This is left as an exercise for you!
A lesssymbolically intense method is to replace "positions" within the relationship with variables and relate them using Pascal's Formula and other known relationships.















Here, we know that a+d=b, c+d=g (or gc=d), d+e=h (or he=d), f+g=j, g+h=k, and h+i=l.
We need to show that (j+k+l)  (a+c+f+b+e+i) = d+g+h. I'll let you work through the arithmetic and the substitutions that lead to success.
1. Consider the letters in the word purchase.
a. How many unique arrangements exist for the letters in the word?
P(8,8) = 8!
b. If an arrangement must begin with a consonant and end with a vowel, how many unique arrangements exist for the letters in the word?
5*3*6! There are 3 consonants to choose from to begin the arrangement and 5 vowels to choose from to end the arrangement. The remaining 6 letters can be arranged in 6! ways.
Exactly 46 credits are required, with 26 of theose already determined in courses in the five areas. This leaves 20 credits yet to be determined. The solution can be found by determining the number of nonnegative integer solutions that exist for the equation M+S+E+H+A = 20, where the letters M, S, E, H, and A respresent the 5 groups.
There are C(20+51,51) = C(24,4) different student programs that could be created.
a. Determine the number of uncollected terms in the expansion.
4^16
b. Determine the number of collected terms in the expansion.
C(19,3)
c. Determine the coefficient P of the collected term PD^6A^2Y^8.
16!/(6!2!8!)
d. How many collected terms in the expansion will have the numerical coefficient P that you determined in (c) above?
The terms will be of the form R^aD^bA^cY^d where a,b,c, and d are from the set {0,2,6,8). There are 4! ways to make this assignment.
First, arrange the three hardback books. There are 3! ways for this to occur. Once these are shelved, there are 4 places among the hardbacks to place the three pairs of paperbacks. These places can be chosen in C(4,3) ways. Finally, there are 6! ways to arrange the six paperback books.
There are 3!*C(4,3)*6! shelving arrangements.