# Number of possible choices

In the experiment of tossing coins, each elementary event represents one of the possible ways for the coin to appear. The number of possible choices for the experiment can be assessed in two different ways.

## Building the sample space

By building the set of all possible elementary events in the experiment called the sample space we can depict all the possible choices as its elements called sample points.

We use H to represent a head and T to represent a tail on the coin.

The total number of possible choices for the experiment is given by the cardinality of the set representing the sample space.

Total number of possible choices for the experiment

 ⇒ n = n(S)
• ### One coin

S = {H, T}  ⇒ n = n(S) = 2
• ### Two coins

S = {HH, HT, TH, TT}

Take the possibilities relating to one coin.

Prefix with H once to get half of the possibilities and then again with T to get the other half.

 ⇒ n = n(S) = 4
• ### Three coins

S = {HHH, HTH, THH, TTH, THH, THT, TTH, TTT}

Take the possibilities relating to one coin.

Prefix with H once to get half of the possibilities and then again with T to get the other half.

 ⇒ n = n(S) = 8

This can be used to build the possibilities for any number of coins.

As the number of coins we consider increase the process of building the set becomes tedious.

## using fundamental counting theorem

For finding the number of possible choices in the experiment, the total event of tossing coins is sub divided into the events of tossing each coin.

The outcomes of the event of tossing multiple coins would be the same as the outcomes of tossing a coin multiple times and collecting the outcomes.

Eg : Tossing a coin 3 times would be the same as tossing a coin thrice.

### Finding Number of possible choices

A coin tossed has two possible outcomes, showing up either a head or a tail.

⇒ The number of possible choices in tossing a coin = 2

• Total Event (E)

The event of tossing the first of the coins

• 1st sub-event (SE1)

The event of tossing the first of the coins

This event can be accomplished in 2 ways

⇒ nSE1 = 2

• 2nd sub-event (SE2)

The event of tossing the second of the coins

This event can be accomplished in 2 ways

⇒ nSE2 = 2

• ...
• ...

Total number of possible choices

 ⇒ nE = nSE1 × nSE2 × .... There will be as many sub events as there are coins. = 2 × 2 × ... = 2(number of coins)

# Number of favorable choices

## Using the sample space

If the sample space is used to specify the total number of possible choices for the experiment, then the number of possible choices for an event would also be found out by building a set of elementary events that are favorable to the occurrence of the event.

A : The event of getting two heads on tossing three coins

A = {HTH, THH, THH}

Number of favorable choices for the event

 ⇒ mA = n(A) = 3

## using permutations and combinations

• Consider the number of heads and tails to appear to make the choice favorable
• Form a word with as many letters as there are heads and tails with H representing heads and T representing tails.
• The number of favorable choices  = Number of arrangement or permutations possible with the word formed Or = Number of combinations with the number of letters in the word form taking either the number of heads or tails at a time

### Example

A : The event of getting four heads on tossing seven coins

Word with H and T representing heads and tails respectively.

HHHHTTT

The order of heads and tails is not important. We just need the word with as many letters as the coins and having as many H's as Heads to appear and as many T's as tails to appear.

Number of favorable choices for the event

= Number of arrangement or permutations with the word HHHHTTT
= Number of permutations with 7 things of which 4 are of a kind and 3 are of another kind and 0 are all different
=
 7! 4! × 3!
 n! a! × b!
=
 7 × 6 × 5 × 4! 4! × 3 × 2 × 1
= 35
Or = Number of combinations of number of letters in the word taking either the number of H's or number of T's at a time
= Number of combinations of 7 things taking 4 at a time
= 7C4
=
 7 × 6 × 5 × 4 4 × 3 × 2 × 1
= 35
Or = Number of combinations of 7 things taking 3 at a time
= 7C3
=
 7 × 6 × 5 3 × 2 × 1
= 35

Using the permutations or combinations route would make it easier to find the number of favorable choices for events involving any number of coins