page23 - Combinations

1) Suppose we wish to arrange 5 people {a, b, c, d, e}, standing side by side, for a portrait. How many such distinct portraits (“permutations”) are possible?
2) {a, b, c, d, e} arranging 5 objects, in such a way that “a” is always first. How many different orders count as being distinct.
3) Now suppose we start with the same 5 people {a, b, c, d, e}, but we wish to make portraits of only k = 3 of them at a time. How many such distinct portraits are possible?
4) Finally suppose that instead of portraits (“permutations”), we wish to form committees (“combinations”) of 3 people from the original 5. How many such distinct committees are possible?
5) Suppose we wish to arrange 5 people {a, b, c, d, e} around a circular table. How many such table arrangements are possible? 

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What is 5! divided by (2! times 3!)?

(5*4*3*2*1)/(2*1*3*2*1) = 10

How many COMBINATIONS of 5 objects taken 2 at a time can be formed?

In how many ways can you PICK 2 out 5 objects to form a group?

How many different HANDFULS of size 2 can you get from 5 objects?

What is 5 CHOOSE 2?

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Choose 2 letters out of these 5: a b c d e

ab, ac, ad, ae, bc, bd, be, cd, ce, de

There are 10 ways to pick 2 out 5 objects

After you have picked 2 out the 5 there are 3 left.

Choosing 3 out of 5 is the same as 2 out of 5 - 10 ways

Obviously picking 1 out of 5 can be done in 5 ways

after picking 1 you are left with 4,
so picking 4 out of 5 - 5 ways

5 out of 5 - 1 way

so NONE out of 5 is - 1 way

Look at the row of Pascal's Triangle: 1, 5, 10 ,10, 5, 1

pick none, pick 1, pick 2, pick 3, pick 4, pick all 5

Virginia Lottery 1992
******Combination Games*******

Price is Right

"Pick a Pair"

6 items pick the two priced the same.

********MORE ON COMBINATIONS************

Suppose the State Lottery takes this form:

6 numbers are randomly chosen from 1 thru 49.

The customer picks 6 numbers.

If the person matches the 6 that were chosen by the State Committee,

they win the entire JACKPOT.

Sometimes this JACKPOT is over $20,000,000.

If the customer was able to buy EVERY POSSIBLE

COMBINATION of 6 numbers, they would have to win.

How many GROUPS of 6 are possible from

the 49 numbers?

In Math we call this 49 CHOOSE 6

or 49 COMBINATION 6

49 objects GROUPED 6 at a time

Some types of word problems are easily solved using Combinations:
Suppose that you are going to choose a small group of 3 items
from a larger group of 7 items.
You could list all the possible groups, if need be.
But, you can count the number of the possible groups
by just using COMBINATIONS.
7 choose 3 has _35__ possible groups

7! / (3! times 4!) = 35

(7*6*5*4*3*2*1*)

(3*2*1)(4*3*2*1)

Reduce by dividing top

and bottom by (4!)

= (7*6*5)/(3*2*1)

OR

LOOKING at PASCAL's TRIANGLESee the row with 

1, 7, 21, 35, 35, 21, 7, 1?

7 choose NONE has _1_ possible group

7 choose 1 has _7_ possible groups

7 choose 2 has _21_ possible groups

7 choose 3 has _35_ possible groups

7 choose 4 has _35_ possible groups

7 choose 5 has _21_ possible groups

7 choose 6 has _7_ possible groups

7 choose 7 has _1_ possible groups

HOW ABOUT using the TI-83?

   

7 choose NONE has _1_ possible group

7 choose 1 has _7_ possible groups

7 choose 2 has _21_ possible groups

7 choose 3 has _35_ possible groups

7 choose 4 has _35_ possible groups

7 choose 5 has _21_ possible groups

7 choose 6 has _7_ possible groups

7 choose 7 has _1_ possible groups

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This forms Sierpiński triangle 

Learn more at:

http://mathyoucancounton.blogspot.com/2013/11/just-for-fun.html

Experiment: flip 3 coins
Outcome: a possible result of the experiment

Sample Space: the set of all possible Outcomes



The Dice Rolling and
Monopoly Videos can be seen
by CLICK HERE.

Click here to see
a video on 8 CHOOSE 2

When viewing if you click on the BOTTOM RIGHT
CORNER the video will FILL THE SCREEN.
To get BACK to the SMALLER VIEW
HIT the "ESC" button on your keyboard.