How many different simple random samples of size 5 can be obtained from a population whose size is 45 ?

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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This value is slightly over 1.2 million

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How I got that

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

with n = 45 and r = 5 to get the same answer in bold above.

$2416.6

step-by-step explanation:

per hour = $ 25

total work time = 35

commission = 15%

total she has to make = $ 4,500

let she sold x dollar products

so her commission will be 15 % of x

now

total = weekly basic pay + commission

4500= (25*35) + 15 % of x

4500=875 + * x

4500-875 =   * x

3625= * x

multiplying wwith 100 on both sides gives

36250 = 15 x

now

x =  

x= $2416.6