Math 3107, Section 2       Friday, May 7

Spring 1999

Suggestions about the homework for Monday, May 10
 
 

Part I: Concerning exercises 11 and 12 (in §11.1)

If you prefer, you may use the formula for the binomial coefficients in the more familiar form:

I'll now explain a method that may be a bit less confusing than what I suggested in class. As a sample, I'll use it to prove the following identity:

.

This example is slightly different from what you're asked to do in either of the exercises, but it's certainly related. Anyway, we begin by working with the expression on the left side of the identity.

Step 1: Cancellation.

After doing the cancellation, we get:

Step 2: Factorization.

In this expression, the numerator always has n as a factor, because:

n! = n(n-1)(n- 2)···2·1 = n[(n-1)(n- 2)···2·1] = n·((n-1)!)

Therefore:

or equivalently:

This last manipulation of the denominator may look like a mere formal game, but in fact it shows that .

Therefore, our calculation proves the identity: .
 
 

Part II: Proving the case n = 10 of exercise 2 in §11 .1.

In the homework for Wednesday, you were asked to guess the value of the expected number of heads when a coin is flipped 10 times. The point of the present exercise is to actually do the calculation, or more specifically, to do the calculation in a way that shows how we could organize similar calculations in the case of a much larger number of coin flips or repetitions of some other experiment. So, it is like exercises 13 and 14 of §11.1, except that we are trying to get the main idea without getting involved in the notational complexities of those exercises!!

Anyway, the formula for the expected number of heads in 10 coin tosses gives a formula like this:

At the beginning, you should explain how you obtained the formula. For one thing, a particular outcome, such as TTHTHTTTTH, has probability . Also, the number of particular outcomes yielding a specific number of heads is given by a binomial coefficient (which one?).

To work with the formula, you certainly can drop the term where the binomial coefficient is multiplied by 0. Then, you can use the identities from exercises 11 and 12 to replace the integer multiples of the binomial coefficients  with multiples of binomial coefficients that have 9 (instead of 10) on top. At this point, a common factor will materialize! After that factor is extracted, the expression inside the large parenthesis will be something for which the binomial theorem will provide considerable simplification (and the final answer).

  A final comment If you want to check your work by just adding up the sum of binomial coefficients, that is certainly OK and probably is a good idea. To receive full credit, however, please be sure to show how the binomial theorem can be used to make everything add up correctly.

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