Sampling Without Replacement
Sampling is called without replacement when a unit is selected at random from the population and it is not returned to the main lot. The first unit is selected out of a population of size $$N$$ and the second unit is selected out of the remaining population of $$N – 1$$ units, and so on. Thus the size of the population decreases as the sample size $$n$$ increases. The sample size $$n$$ cannot exceed the population size $$N$$. Once the unit is selected for a sample it cannot be repeated in the same sample. Thus all the units of the sample are distinct from one another. A sample without replacement can be selected either by using the idea of permutations or combinations. Depending upon the situation, we write all possible permutations or combinations. If the different arrangements of the units are to be considered, then the permutations (arrangements) are written to get all possible samples. If the arrangement of units is of no interest, we write the combinations to get all possible samples.
Combination
Let us again consider a lot (population) of $$5$$ bulbs with $$3$$ good ($${G_1},{G_2}$$and$${G_3}$$) and 2 defective ($${D_1}$$and$${D_2}$$) bulbs. Suppose we have to select two bulbs in any order. There are $$^5{C_2} = \frac{{5!}}{{2!3!}} = 10$$ possible combinations or samples. These combinations (samples) are listed as: $${G_1}{G_2}$$, $${G_1}{G_3}$$, $${G_2}{G_3}$$, $${G_1}{D_1}$$, $${G_1}{D_2}$$, $${G_2}{D_1}$$, $${G_2}{D_2}$$, $${G_3}{D_1}$$, $${G_3}{D_2}$$, $${D_1}{D_2}$$.
There are $$10$$ possible samples and each of them has a probability of selection equal to $$1/10$$. The selected sample will be any one of these $$10$$ samples. The sample selected in this manner is also called a simple random sample. In general, the number of samples by combinations is equal to \[^N{C_n} = \frac{{N!}}{{n!\left( {N – n} \right)!}}\].
Permutation
Each combination generates a number of arrangements (permutations). Thus in general the number of permutations is greater than the number of combinations. In the previous example of bulbs, if the order of the selected bulbs is to be considered then the number of samples by permutation is given by \[^5{P_2} = \frac{{5!}}{{\left( {5 – 2} \right)!}} = 20\].
These samples are:
$${G_1}{G_2}$$ 
$${G_2}{G_1}$$

$${G_1}{G_3}$$

$${G_3}{G_1}$$

$${G_2}{G_3}$$

$${G_3}{G_2}$$

$${G_1}{D_1}$$

$${D_1}{G_1}$$

$${G_1}{D_2}$$

$${D_2}{G_1}$$

$${G_2}{D_1}$$

$${D_1}{G_2}$$

$${G_2}{D_2}$$

$${D_2}{G_2}$$

$${G_3}{D_1}$$

$${D_1}{G_3}$$

$${G_3}{D_2}$$

$${D_2}{G_3}$$

$${D_1}{D_2}$$

$${D_2}{D_1}$$

Each sample has a probability of selection equal to $$1/20$$. The selected sample maintains that the order of the bulbs will be any one of these $$20$$ samples. A sample selected in this manner is also called a simple random sample because each sample has an equal probability of being selected.