This value is slightly over 1.2 million
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How I got that answer:
We have 45*44*43*42*41 = 146,611,080 different permutations. I started with 45 and counted down by 1 until I had 5 values multiplied out.
If order mattered, then we would stop here. But order doesn't matter. All that matters is the group overall rather than any particular individual or how they rank in the group.
There are 5! = 5*4*3*2*1 = 120 different ways to arrange a five item set, so the very large result we got earlier is overcounting by a factor of 120.
To fix this, we divide by 120 to get
(146,611,080)/120 = 1,221,759
This represents the number of combinations in which we're selecting 5 members from a pool of 45 total.
You could use the nCr formula
[tex]_nC_r = \frac{n!}{r!*(n-r)!}[/tex]
with n = 45 and r = 5 to get the same answer in bold above.
Using the principle of Combination, the number of different random samples of size five which can be obtained form a population of 45 is 1221759
Using the principle of combination, which is defined as :
Hence,
45C5 = 45! ÷ (45 - 5)!5!
45C5 = 45! ÷ 40!5!
45C5 = (45 × 44 × 43 × 42 × 41) / (5 × 4 × 3 × 2 × 1)
45C5 = 1221759
Therefore, there are 1221759 ways of making a selection of 5 samples from a population of size 45.
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