a) Arrangements containing 5 different letters from the word AMPLITUDE are to be made.
i) the number of 5-letter arrangements if there are no restrictions,
Since there are 9 different letters, and we pick 5 to be arranged, there are 9P5 = 15,120 permutations.
ii) the number of 5-letter arrangements which start with the letter A and end with the letter E.
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 7P3 = 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.
This means selecting 15 students from 20, so we have 20C15 = 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,
Selecting 10 boys from 12, we have 12C10 = 66 ways.
Selecting 5 girls from 8, we have 8C5 = 56 ways.
∴ the total is 12C10 × 8C5 = 3,696 ways.
iii) all the boys get tickets.
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 8C3 = 56 ways.
∴ the total is 8C3 × 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.
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.
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.
Selecting 4 seniors from 6, we have 6C4 = 15 selections.
Selecting 2 juniors from 5, we have 5C2 = 10 selections.
∴ the total is 6C4 × 5C2 = 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.
"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 5C3 × 4C1 = 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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