Chapter 7

Permutation and Combinations

Fundamental Principle of Counting:

If one experiment has n possible outcomes and another experiment has m possible outcomes, then there are m × n possible outcomes when both of these experiments are performed simultaneously. 

In other words, suppose a job has n parts and the job will be completed only when each part is completed. Further, it is known that the first part can be completed in a1 ways, the second part can be completed in a2 ways and so on ...... the nth part can be completed in an ways. Then the total number of ways of doing the job is a1a2a3 ... an. This is also known as the rule of product.

Note:

  • Fundamental Principal of Counting (FPC) is used to calculate the possibilities of an event without actually counting them.
  • Basic steps to be used while applying the FPC:

1. Try to identify the independent events involved in a given problem.

2. Find the number of ways of performing or possibilities of occurrence of an event.

3. Multiply these numbers to get the total number of ways of occurrence of these events.

Example 1:

Suppose you have 33 shirts (call them A , B , and C ), and 44 pairs of pants (call them w , x , y , and z ).

Ans.: Then you have

3×4=12

possible outfits:

AwAxAyAz, BwBxByBz, CwCxCyCz

Example 2:

Suppose you roll a 6 -sided die and draw a card from a deck of 52 cards. There are 6 possible outcomes on the die, and 52 possible outcomes from the deck of cards.

Ans.:

So, there are a total of

6×52=312

possible outcomes of the experiment.

Generalisation of the Addition and the Product Rule

In general, if there are several mutually exclusive events P1, P2, P3, P4……Pn…etc. with the respective number of ways given as n (P1), n(P2), n(P3), n(P4)….n(Pn),  then the number of ways in which either P1 and P2 and ….. … Pn can occur is given by,

n(E) = n(P1) + n(P2) …….. + n(Pn)

Similarly, if there are several mutually independent events P1, P2, P3, P4……Pn…etc. with the respective number of ways given as n (P1), n(P2), n(P3), n(P4)….n(Pn), then the number of ways in which  P1 and P2 and ….. … Pn  can occur is given by,

n(E) = n(P1) × n(P2) …….. × n(Pn)

We must note that all the possible number of ways derived thus, all of them will represent the unique and distinct ways in which the event E will take place.

Event and Its Trials

Suppose we have an event E with ‘m’ possible outcomes. If we conduct this event n number of times, then the number of outcomes of ‘n trials of the event’ will be, mn

Clearly, this is only valid when all the outcomes of the experiment/event E are independent of each other. This would ensure that everytime the event takes place, any of its outcomes are possible.

Question : A coin is tossed 50 times. What is the number of possible outcomes of this experiment?

Answer : A coin toss has two possible outcomes: ‘A heads’ or ‘A tails’. Every toss of the coin is independent of every other toss of the coin, since whenever you toss the coin; there would be two possible outcomes with equal probability. Thus, the no. of possible outcomes of the experiment = 250

Question :

How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that

(i) repetition of the digits is allowed?

(ii) repetition of the digits is not allowed?

ANSWER:

(i)

There will be as many ways as there are ways of filling 3 vacant places

https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/4203/chapter%207_html_m70fc0d6.jpghttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/4203/chapter%207_html_m70fc0d6.jpghttps://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/4203/chapter%207_html_m70fc0d6.jpg in succession by the given five digits. In this case, repetition of digits is allowed. Therefore, the units place can be filled in by any of the given five digits. Similarly, tens and hundreds digits can be filled in by any of the given five digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed from the given digits is 5 × 5 × 5 = 125

(ii)

In this case, repetition of digits is not allowed. Here, if units place is filled in first, then it can be filled by any of the given five digits. Therefore, the number of ways of filling the units place of the three-digit number is 5.

Then, the tens place can be filled with any of the remaining four digits and the hundreds place can be filled with any of the remaining three digits.

Thus, by the multiplication principle, the number of ways in which three-digit numbers can be formed without repeating the given digits is 5 × 4 × 3 = 60

Question :

A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?

ANSWER:

When a coin is tossed once, the number of outcomes is 2 (Head and tail) i.e., in each throw, the number of ways of showing a different face is 2.

Thus, by multiplication principle, the required number of possible outcomes is 2 × 2 × 2 = 8