Can somone please help me to solve the following
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Can somone please help me to solve the following

[From: ] [author: ] [Date: 11-11-02] [Hit: ]
/(3!x2!).5. 5C5 = 5!/(5!......
{1} 11C10

{2} 5p5

{3} 6!

{4} 5C2

{5} 5C5

{6} 5C3 over 5! it's written as a fraction

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1. 11C10=11C1=11. explaination: 11C10 means" number of ways to choose 10 items from 11 items" which is the same as "number of ways choose 1 item to leave behind from 11 items" which is 11C1 which intuitively is 11.

2.5p5 = 5!= 5x4x3x2x1.

3. 6! = 6x5x4x3x2x1.

4. 5C2 = 5!/(3!x2!).

5. 5C5 = 5!/(5!x0!) = 1.

6. 5C3/5! = 5!/(3!x2!x5!) = 1/(3!x2!).

Please understand what ! means:
3! = 3x2x1.
2! = 2x1.
1! = 1.
0! = 1 (unusual case).
I hope you get the general idea here. Also type to understand the C symbol. It gives number of ways to chose a subset from a set and the formula is 5C3 =5!/(3!2!).
The p symbol is a permute function: 5p3 gives number of ways to choose and then arrange 3 items from 5 items.
You can find all these functions on your calculator.
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