Get better study experience with the jambite for andriod app!
• Take Past Questions Offline
• Enjoy our online tutorial classes offline
• use the jambite app at your convenience anywhere anytime! ok{[seenAndroidpopup]}
• # [ended] Mathematics Tutoring Session 3 (Permutation and Combination Lesson [Fri 19th Feb 2016, 4:00 - 5:00pm])

Mathematics

Emmanuel Iwara(Tutor)
19-02-2016 13:23:00 +0000

The concept of PERMUTATION is often tested on jamb and other exams. Understanding this topic is critical to passing the jamb mathematics exams.

FACTORIALS

Definiton

The product of the first n natural numbers is called factorial n and denoted by n!

n! = n(n-1)(n-2)(n-3)(n-4)... (4)(3)(2)(1).

That is 4! = 4 x 3 x 2 x 1 = 24
3! = 3 x 2 x 1 = 6

In general,
There are precisely n! different ways of arranging n objects in order.

Example 1. In how many different ways can a committee of 8 people be seated at a table if the table is the usual rectangular shape with the president at one end.

Soln
The seats can be put in order. The president's seat can be regarded as the first, etc. We than have 8 poeple to be arranged on 8 seats in order.

= 8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40320

Note; The number of permutations of n distinct objects arranged in a circle is (n-1)!

PERMUTATION
Definition

The number of different ways of arranging r objects in order selected from a group of n is called the number of permutation of r from n is denoted by n! / (n-r)!
Therefore there are n! / (n-r)!
Permutations of r object chosen from n unlike objects, i r is less than n.

Example 1. In how many different ways can 3 books be put into 6 bags not more than one in each?

Soln
Number of ways =6P3
= 6! / (6-3)!
= 6! / (3)!
= 6 x 5 x 4 x 3 x 2 x 1 / 3 x 2 x 1
= 720 / 6 = 120

The number of distinct permutation of n objects of which r1 are of one kind, r2 of a second kind,...nk
n! / r1!,r2!, ...rk

Example 2. How many permutations can be made from the letters of the word EMMANUEL.

Soln

The name is of eight letters with M and E repeated two times
The number of permutations is
8! / 2! x 2!
8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 / 2 x 1 x 2 x 1
40320 / 4
= 10080

Now see if you can try these;

(a) How many permutations can be made from the letters of the word MISSISSIPPI.

(b) How many different 3-digit numbers can be formed formed from the digits 0, 1, 2, 3 if none may be used more than once?
(c) How many odd numbers can be formed by using all the digits 1,2,3,4,5?

(d) In how many ways can the letters of the word STEMMAR be arranged if the two M's must not come together?

**Note that this session begins from 4:00 - 5:00pm. Before the session begins, you can put down your questions on paper and then ask them when the session officially begins by 4:00pm.

This session has ended

Thanks to everyone that participated.

• Joshua Musa

okay

0 19-02-2016 13:29:00 +0000

• Teslimat Umoru

Cool

0 19-02-2016 13:43:00 +0000

• Cisco Ramon

Thanks

0 19-02-2016 13:47:00 +0000

• Akande Jamiu

(a) 8!4!4!x2! =8x7x6x5x4x3x2x1x4x3x2x1x4x3x2x1x2x1= 46448640

0 19-02-2016 13:50:00 +0000

• Akande Jamiu

(b) :- 3!(3-3)!= 3!0! = 3!= 3x2x1 =6

0 19-02-2016 13:57:00 +0000

0 19-02-2016 13:59:00 +0000

• Akande Jamiu

(d) 7!5!=7x6x5x4x3x2x1x5x4x3x2x1 = 604800

0 19-02-2016 14:31:00 +0000

• asmau yusuf

Looking forward to it

0 19-02-2016 14:41:00 +0000

• Akande Jamiu

(c) 5!2!=5x4x3x2x1x2x1 = 240

0 19-02-2016 14:43:00 +0000

• Odudele Paul

Factorise a^3+(3b)^3

0 19-02-2016 14:44:00 +0000