Let n be a positive integer. (a) prove that n^3 = n + 3n(n - 1) + 6 c(n, 3) by counting the number of ordered triples (a, b,c), where 1 < = a, b, c < = n, in two different ways. (b) prove that c(n + 2, 3) = (1)(n) + (2)(n - 1) + (3)(n - 2) + . . + (k)(n - k + 1) + . . + (n)(1), by counting the number of subsets of {1, 2, 3, . ., n + 2} containing three different numbers in two different ways.

(a) Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

(b) To choose three numbers, we can choose two groups one group with two numbers and the other group has one number.  The total of n + 2 numbers can be separated into two groups.

In the first way, one group has 2 numbers, and the other group has n numbers.  They form a total of n + 2 numbers.  There are C(2,2) = 1 ways to choose two numbers from the 2 group.  There are C(n,1) = n ways to choose one number from the one group.  This gives us a first term of 1*n.

In the second way, one group has three numbers, and the other group has n - 1 numbers.  They form a total of n - 1 numbers.  There are C(3,2) = 2 ways to choose two numbers from the 2 group.  There are C(n - 1,1) = n - 1 ways to choose one number from the one group.  This gives us a second term of 2*(n - 1).  The pattern will continue until we reach n.  So the two sides are equal.

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

you forgot the options, but i found them anyways, the correct answer is "john wilson believes that words and actions based on the intellect are superior to those based on emotions".

john wilson is depicted as an extremely intelligent man. the point of the excerpt is to make a sort of distinction between emotional and rational intelligence. he happens to have both attributes, but he clearly favors his intellectual gifts. at the same time, because of his religious background, he's a little bit conflicted with this and it makes him ashamed.

hope this !