Charlie reads quickly. He reads 1\dfrac371 7 3 1, start fraction, 3, divided by, 7, end fraction pages every \dfrac23 3 2 st
2 answers:
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Answer:
a) 20 ways
b) 10 ways
Step-by-step explanation:
When the order of selection/choice matters, we use Permutations to find the number of ways and if the order of selection/choice does not matter, we use Combinations to find the number of ways.
Part a)
We have to chose 2 objects from a group of 5 objects and order of choice matters. This is a problem of permutations, so we have to find 5P2
General formula of permutations of n objects taken r at time is:
Using the value of n=5 and r=2, we get:
Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice matters.
Part b)
Order of choice does not matter in this case, so we will use combinations to find the number of ways of choosing 2 objects from a group of 5 objects which is represented by 5C2.
The general formula of combinations of n objects taken r at a time is:
Using the value of n=5 and r=2, we get:
Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice does not matters.
<h2>Step-by-step explanation:</h2><h3>
Given</h3>
s10 = -20
n = 10
an = -20
solution
s10 = n\2 [a+(n-1)d]
-20 = 10a2[a+(10-1)4]
-4 = a +36
a1 = -40
Hope it helps.
Hope it helps