ESSENTIAL COUNTING RULES

Example: A certain two-symbol code consists of a letter in the first position and a digit in the second position.

Give some elements of the sample space. S.S. =

{B3, A7, Z0,

X7 . . . }How many possible codes are there?

0 1 … 8 9

A

B

…

Z

A0 A1 … A8 A9

B0 B1 … B8 B9

Z0 Z1 … Z8 Z9

All in all, we have 26x10 = 260 possible codes

General Principle

If one event can occur in m ways and a second event can occur in n ways, then all-in-all the number of ways both events can occur is m x n. (Same applies for more than two events.)

Example: In a medical study, patients are classified according to blood type (AB+, AB-, A+, A-, B+, B-, O+, O-) and also according to blood pressure (low, normal, high).

S.S. = { (AB+,low) (A+,high) (B-,high) … }

Give some elements of the sample space.

How many possible classifications are possible? 8 x 3 = 24 classifications

Example: A box contains 26 balls each one labelled with a letter, and three balls are taken in succession.

S.S. = { CAY ZXT THX … }

Give some elements of the sample space.

How many possible triples are possible? 26 x 25 x 24 = 15600 triples

This is what we call a permutation. Permutations are the no. of ways in which a set of objects can be arranged in order.

Example: Give a few elements of the sample space first.

1. If repetitions are not permitted, how many 3 digit numbers can be formed from the six digits 2, 3, 5, 6, 7 and 9? (ii) how many of these are less than 400?

(iii) how many are even?

(iv) how many are odd?

(iv) how many are multiples of 5?

3. In how many ways can 3 boys and 2 girls sit in a row?

(ii) In how many ways can they sit in a row if the boys and girls are each to sit together?

(iii) In how many ways can they sit in a row if just the girls are to sit together?

2. Do #1, if repetitions are permitted.

4. Find the number of four letter words that can be formed from the letters of the word HISTORY.

(i) How many of them contain only consonants? (ii) How many of them begin and end in a consonant

(iii) How many of them begin with a vowel? (iv) How many contain the letter Y?

(v) How many begin with T and end in a vowel?

(vi) How many begin with T and also contain S? (vii) How many contain both vowels?

In contrast to permutations, combinations are the no. of ways in which a set of objects without regard to ordering.

Example: A box contains 7 balls each one labelled A to G, and three balls are taken.

If three balls are taken in succession, then the no. of all possible triples is:7 x 6 x 5 = 210 (permutations)

In this particular process of selection, for example:

ZTX is not the same as ZXT, XZT, XTZ, TZX, TXZ (6 = 3!)

QSB is not the same as QBS, SQB, SBQ, BQS, BSQ (6 = 3!)

LUV is not the same as LVU, UVL, ULV, VLU, VUL (6 = 3!)

And so forth.

How about if three balls are taken at once?

In this case, the ordering no longer matters, for example:

In this particular process of selection, for example:

ZTX, ZXT, XZT, XTZ, TZX, TXZ count as one (6 = 3!)

QSB, QBS, SQB, SBQ, BQS, BSQ count as one (6 = 3!)

LUV, LVU, UVL, ULV, VLU, VUL count as one (6 = 3!)

And so forth.

Again, this is what we call combinations. These combinations divide the unique permutations (7x6x5=210) into groups of 6 (=3!) triples that count as one.

The total no. of combinations in this case is:210/6 = 35

7

3

7

3is read as: combination of 7 objects taken 3 at once

In a short symbol, we write:

7!3! (7 3)!

35

The number of combinations of n objects taken r at a time (i.e., a selection of r objects from a set of n objects regard-less of ordering) is,

!

! !

n nk k n k

Example: How many groups consisting of 3 people can be formed from 5 people?

We have a total of 5 people: say, A, B, C, D, E

And we want to select committees of 3 people from this set regardless of order.

Example: ABC (same as ACB, BAC, BCA, CAB and CBA) BDE (same as BED, DBE, DEB, EBD and EDB)

Therefore, the no. of committees is:

5 5!10

3!2!3

Example: A delegation of 4 students is selected each year from a college to attend a conference. (i) In how many ways can the delegation be chosen if there are 12 eligible students? (ii) In how many ways if two of the eligible students will not attend the meeting together? (iii) In how many ways if two of the eligible students are married and will only attend the meeting together?

Example: A student is to answer 8 out of 10 questions on an exam. (i) How many choices has he? (ii) How many if he must answer the first 3 questions?

Example: In how many ways can a committee consisting of 3 men and 2 women be chosen from 7 men and 5 women?