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About LinkBacksThis is a combinations problem. Because all we want is the number of different groups of people not which one is first or second, thus,Originally Posted by Kiwigirl
!}=\frac{19!}{12!\cdot 7!}=50388)
1. A lift in a building can carry a maximum of 12 passengers. 19 people are waiting for the lift. In how many ways can the passengers be carried in two journeys?
2. How many groups of 4 letters can be formed from the letters Y, S, L, K, I, if the second letter chosen must always be I? (Letters cannot be repeated.)
3. Five chocolates are selected at random from a box. In the box there are 5 strawberry creams, 7 orange chip, 7 toffee choc and 8 nut crunch.
a) How many different combinations of five chocolates can be chosen?
b)Tara would like 2 orange chip, 2 toffee choc and 1 strawberry cream. What is the probability she chooses this selection?
You are chosing 4 letters thus you have,
_ _ _ _
Now the second one must be an 'I' thus, you have,
_ I _ _
For the first letter you can chose only 4 different possibilities because you already used 'I'. For the third letter you can chose only 3 becauase you already used an 'I' and the first letter cannot be repeated. And for the last you only have 2 possibilities. Thus, you have,
a) You can have, let S,O,T,N represent what all the number of chocolates you are taking.
Thus,
Where
5,0,0,0 x4
4,1,0,0 x12
3,1,1,0 x12
3,2,0,0 x12
2,1,1,1 x4
2,2,1,0 x12
TOTAL=56 possibilities.
Let me just explain what I did, you can express 5 as 5+0+0+0 then you need to find distinct similiar representations such as 0+5+0+0 or 0+0+5+0 or 0+0+0+5, thus a total of 4 (that is what x4 means).
b)To find the probability you need to find all possible selections of 2 Orange 2 Toffee and 1 Strawberry. And divide it by all possible 5 number selections. The number of favorable selections (2O 2T 1S) iswhich is equal to 105. There are
. Thus the probability is
(extremely low).
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