There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

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

Thus the formula for this particular combination is,

\(\displaystyle C (8,5) = P \frac{(8,5)}{5!}\)

\(\displaystyle \frac{8\times 7\times 6\times 5\times 4}{5\times 4\times 3\times 2\times 1} =\)

The \(\displaystyle 5 \times 4\) will cancel out because it is in the numerator and denominator,

\(\displaystyle \frac{8\times 7\times 6}{3\times 2\times 1} = \frac{336}{6} = 56\).

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

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

Thus the formula for this particular combination is,

\(\displaystyle C (10,4) = P \frac{(10,4)}{4!}\)

\(\displaystyle \frac{10\times 9\times 8\times 7}{ 4\times 3\times 2\times 1} =\)

\(\displaystyle \frac{90\times 56}{12\times 2} = \frac{5040}{24} = 210\)

There are two types of statistical calculations that are used when dealing with ordering a number of objects. When the order does not matter it is known as a combination and denoted by a C.

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

Thus the formula for this particular combination is,

\(\displaystyle C=\frac{P(6,2)}{2!}\) 

There are 2 points on each line segment.

 \(\displaystyle \frac{6\times 5}{ 2\times 1} = \frac{30}{2} = 15\)

Evaluate \(\displaystyle _{5}C_{3}\) is asking to calculate the combination of five objects when choosing three of them.

\(\displaystyle _{5}C_{3} = \frac{5!}{(5-3)! 3!}\) 

\(\displaystyle _{5}C_{3} = \frac{5!}{(5-3)! 3!} = \frac{5\times 4\times 3\times 2\times 1}{2\times 1\times3 \times 2\times 1}\)

\(\displaystyle 3!\) or \(\displaystyle 3\times2\times1\) cancels out.

\(\displaystyle _{5}C_{3} = \frac{5\times 4 }{2\times 1}\)

\(\displaystyle \frac{20}{2} = 10\)

Step 1: Count how many numbers I can use..

I can use 9 numbers.

Step 2: Determine how many numbers I can put in the first digit of the three-digit number..

I can put \(\displaystyle 8\) numbers in the first spot. I cannot put \(\displaystyle 0\) in the first slot because the number will not be a three-digit number.

Step 3: Determine how many numbers I can put in the second digit..

I can also put \(\displaystyle 8\) numbers in the second spot. Here's the reason why it's still \(\displaystyle 8\):

Let's say I choose 2 for the first number. I will take \(\displaystyle 2\) out of my set. I had \(\displaystyle 9\)numbers in my set..If i take a number out, I still have \(\displaystyle 8\) numbers left. These numbers are: \(\displaystyle 3,4,5,6,7,8,9,0\).

Step 4: Determine how many numbers I can put in the third and final digit...

I can put \(\displaystyle 7\) numbers in the third slot..

I had \(\displaystyle 9\) numbers at the start, and then I removed \(\displaystyle 2\) of them. \(\displaystyle 9-2=7\).

Step 5: Multiply how many numbers can go in the first, second, and third spot..

\(\displaystyle 8\cdot8\cdot7=448\).

There are a total of \(\displaystyle 448\) non-repetitive three-digit numbers that can be formed. 

Step 1: Recall the combination formula...

\(\displaystyle _nC_r=\frac {n!}{r!(n-r)!}\)

Step 2: Find \(\displaystyle n\) and \(\displaystyle r\) from the question..

\(\displaystyle n=17, r=9\).

Step 3: Plug in the values into the formula above..

\(\displaystyle \frac {17!}{9!(17-9)!}=\frac {17!}{9!\times8!}=24,310\)

This problem is solved by knowing that we have six options for first place, five options for second place, and so on.

Which means 

\(\displaystyle 6*5*4*3*2*1 = 720\)