Worked examples of problems on Permutations & Combinations as typically found in
CIE IGCSE syllabus 0606 (and also CIE GCE O-Level syllabus 4037) - Additional Mathematics Paper 1 May/June 2012.

Additional Maths Paper 1 May/June 2012 (pdf)

The following figure gives the formula for Permutations and Combinations. Scroll down the page for examples and solutions on how to use the formulas to solve examination word problems.

### 0606 S12 Paper 11 Question 4

### 0606 W12 Paper 21 Question 9

You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

Additional Maths Paper 1 May/June 2012 (pdf)

The following figure gives the formula for Permutations and Combinations. Scroll down the page for examples and solutions on how to use the formulas to solve examination word problems.

a) Arrangements containing 5 different letters from the word AMPLITUDE are to be made.

Find

i) the number of 5-letter arrangements if there are no restrictions,

Solution:

Since there are 9 different letters, and we pick 5 to be arranged, there are ^{9}P_{5} = 15,120 permutations.

ii) the number of 5-letter arrangements which start with the letter A and end with the letter E.

Solution:

Since A and E are fixed, there are only 3 other letters to arrange in between them, from the remaining 7 letters (9 letters minus the A and E).

∴ there are ^{7}P_{3} = 210 permutations.

b) Tickets for a concert are given out randomly to a class containing 20 students. No student is given more than one ticket. There are 15 tickets.

i) Find the number of ways in which this can be done.

Solution:

This means selecting 15 students from 20, so we have ^{20}C_{15} = 15,504 ways.

There are 12 boys and 8 girls in the class. Find the number of different ways in which

ii) 10 boys and 5 girls get tickets,

Solution:

Selecting 10 boys from 12, we have ^{12}C_{10} = 66 ways.

Selecting 5 girls from 8, we have ^{8}C_{5} = 56 ways.

∴ the total is ^{12}C_{10} × ^{8}C_{5} = 3,696 ways.

iii) all the boys get tickets.

Solution:

All 12 boys got tickets, so there is only 1 way to select all the boys.

The remaining 3 tickets go to 3 girls from 8, we have ^{8}C_{3} = 56 ways.

∴ the total is ^{8}C_{3} × 1 = 56 ways.

a) An art gallery displays 10 paintings in a row. Of these paintings, 5 are by Picasso, 4 by Monet and 1 by Turner.

i) Find the number of different ways the paintings can be displayed if there are no restrictions.

Solution:

Since there are 10 different items to be arranged, there are 10! = 3,628,800 permutations.

ii) Find the number of different ways the paintings can be displayed if the paintings by each of the artists are kept together.

Solution:

Picasso's paintings can be arranged in 5! ways,

Monet's can be arranged in 4! ways and

Turner's can be arranged in 1! = 1 way.

Also, the 3 artists can be arranged in 3! ways (P-M-T, P-T-M, M-T-P, etc)

∴ the total is 5! × 4! × 3! = 17,280 permutations.

b) A committee of 4 senior students and 2 junior students is to be selected from a group of 6 senior students and 5 junior students.

i) Calculate the number of different committees which can be selected.

Solution:

Selecting 4 seniors from 6, we have ^{6}C_{4} = 15 selections.

Selecting 2 juniors from 5, we have ^{5}C_{2} = 10 selections.

∴ the total is ^{6}C_{4} × ^{5}C_{2} = 150 selections.

One of the 6 senior students is a cousin of one of the 5 junior students.

ii) Calculate the number of different committees which can be selected if at most one of these cousins is included.

Solution:

"At most one" means that there may be none of the cousins, or only one of the cousins included. So
we can work out all the different scenarios: none of the cousins, only the senior cousin included, or only the junior cousin included and add all the selections.

An alternative is to work out the number of selections where both cousins are in; then we subtract from the total (from b(i) above) and the remaining
selections would have at most one of the cousins.

If both cousins are included, then we select only 3 other seniors from the remaining 6, and 1 other junior from the remaining 4.

So, there are ^{5}C_{3} × ^{4}C_{1} = 40 selections where both cousins are included.

∴ there are 150 − 40 = 110 selections where there is at most one of the cousins.

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