Question

1. Justification of the two formulas for permutation and combinations. (15 pts) a) Assume there are n different objects and we want to place them in k different places. By drawing and simplifying, justify that there are P(n,k) ways. P b) Now assume that places are indistinguishable. Use item a to show then answer is C(n,k) as it follows. Explain your answer why do we have to divide by k!. ( n! kn ethe binomial theorem: (a + b)" = Σ() a n k n-k c) Using item b) (combination) prove the binomial theorem: (a + b Where "a, b, n " are constants. Hint: write (a+b)n as (a+b)(a+b(a+b) "n times" and use logic and counting techniques.

Answer 1

a) Consider different places.

The first place can be filled in ways.

The second place can be filled in ways.

The third place can be filled in ways.

n-2

.........................................................................

The k-th place can be filled in ways.

Together the number of ways is $n\left (n-1 \right )...\left (n-k+1 \right )$ ways.

ways.

n!

This is called $P(n,k)=\frac{n!}{\left ( n-k \right )!}$

b) When the places are indistinguishable, the number reduces by a factor of . Because places can be ordered in ways. That is the required ways is

ways.

(n) = An,k) = k! (n-k)!

c) Here $(a+b)^n=(a+b)(a+b)...(a+b)$ . number of a's can be selected in $\binom{n}{k}=\frac{n!}{k!\left ( n-k \right )!}$ ways. So the coefficient of
akbn
is $\binom{n}{k}=\frac{n!}{k!\left ( n-k \right )!}$ .

Thus,

$(a+b)^n=(a+b)(a+b)...(a+b)=\sum_{k=0}^{n}\frac{n!}{k!\left ( n-k \right )!}a^kb^{n-k}$

The proof is complete.