Question

Let C(n, k) = the number of k-membered subsets of an n-membered set. Find (a) C(6, k) for k = 0,1,2,...,6 (b) C(7, k) for k = 0,1.2...,7

Answer 1

Answer:

(a) $C(6,0) = 1$, $C(6,1) = 6$, $C(6,2) = 15$, $C(6,3) = 20$, $C(6,4) = 15$, $C(6,5) = 6$, $C(6,6) = 1$.

(b) $C(7,0) = 1$, $C(7,1) = 7$, $C(7,2) = 21$, $C(7,3) = 35$, $C(7,4) = 35$, $C(7,5) = 21$, $C(7,6) = 7$, $C(7,7)=1$.

Step-by-step explanation:

In this exercise we only need to recall the formula for C(n,k):

$C(n,k) = \frac{n!}{k!(n-k)!}$

where the symbol $n!$ is the factorial and means

$n! = 1\cdot 2\cdot 3\cdot 4\cdtos (n-1)\cdot n$.

By convention 0!=1. The most important property of the factorial is $n!=(n-1)!\cdot n$, for example 3!=1*2*3=6.

(a) The explanations to the solutions is just the calculations.

• $C(6,0) = \frac{6!}{0!(6-0)!} = \frac{6!}{6!} = 1$

• $C(6,1) = \frac{6!}{1!(6-1)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

• $C(6,2) = \frac{6!}{2!(6-2)!} = \frac{6!}{2\cdot 4!} = \frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

• $C(6,3) = \frac{6!}{3!(6-3)!} = \frac{6!}{3!\cdot 3!} = \frac{5!\cdot 6}{6\cdot 6} = \frac{5!}{6} = \frac{120}{6} = 20$

• $C(6,4) = \frac{6!}{4!(6-4)!} = \frac{6!}{4!\cdot 2!} = frac{5!\cdot 6}{2\cdot 4!} = \frac{4!\cdot 5\cdot 6}{2\cdot 4!} = \frac{5\cdot 6}{2} = 15$

• $C(6,5) = \frac{6!}{5!(6-5)!} = \frac{6!}{5!} = \frac{5!\cdot 6}{5!} = 6$

• $C(6,6) = \frac{6!}{6!(6-6)!} = \frac{6!}{6!} = 1$.

(b) The explanations to the solutions is just the calculations.

• $C(7,0) = \frac{7!}{0!(7-0)!} = \frac{7!}{7!} = 1$

• $C(7,1) = \frac{7!}{1!(7-1)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

• $C(7,2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2\cdot 5!} = \frac{6!\cdot 7}{2\cdot 5!} = \frac{5!\cdot 6\cdot 7}{2\cdot 5!} = \frac{6\cdot 7}{2} = 21$

• $C(7,3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!\cdot 4!} = \frac{6!\cdot 7}{6\cdot 4!} = \frac{5!\cdot 6\cdot 7}{6\cdot 4!} = \frac{120\cdot 7}{24} = 35$

• $C(7,4) = \frac{7!}{4!(7-4)!} = \frac{6!\cdot 7}{4!\cdot 3!} = frac{5!\cdot 6\cdot 7}{4!\cdot 6} = \frac{120\cdot 7}{24} = 35$

• $C(7,5) = \frac{7!}{5!(7-2)!} = \frac{7!}{5!\cdot 2!} = 21$

• $C(7,6) = \frac{7!}{6!(7-6)!} = \frac{7!}{6!} = \frac{6!\cdot 7}{6!} = 7$

• $C(7,7) = \frac{7!}{7!(7-7)!} = \frac{7!}{7!} = 1$

For all the calculations just recall that 4! =24 and 5!=120.