Counting the Number of Samples Using Tree Diagram and Formula, Statistical Note 28

How many samples of size ‘r’ can be drawn from the population of size ‘n’?

All may know that a finite population can have many samples of equal size, but how many?The number of samples than can be drawn from a finite population varies whether the sampling units are drawn without or with replacement. I discussed it in my previous Statistical Note 15 (Counting of Outcomes) using tree diagram and formula to different examples. In this note, I take only two examples six and eight discussing on using tree diagram and formula to draw some balls from the population of balls without or with replacement in which the order of the ball does not matter. Most often the position of the sampling units in a sample does not matter. The number of samples will be important to discuss on sampling distribution in future statistical notes.

Sampling Without Replacement

A bag contains four balls of different colors (A of Red color, B Blue, C Green and D yellow). Select two balls randomly one at a time and do not replace back to the bag after selecting. Calculate the total number of outcomes containing two balls selected without replacement in which the order of the balls does not matter.

Tree Diagram: This example has six combinations of 12 outcomes each constituting two balls selected one at a time without replacement from the total of four balls, Red (A), Blue (B), Green (C) and Yellow (D) in which the order of balls does not matter. The outcomes are indicated by numbers on the right most part of the Diagram 1.

Diagram 1: Outcomes in Selecting Two Out of Four Balls Without Replacement in which the Order of Balls does not Matters

Formula: In the above example, the combination formula n!/(n-r)!*r! is used (see Diagram  1 of Statistical Note 15). In this case, ‘n’ is equal to 4 and ‘r’ equal to 2 balls. Using these values in the formula, we get 4!/(4-2)!*2!, which is equal to 4!/2!*2!, equal to 6. The combination formula is denoted by C(n,r) or ⁿC_r also.

Thus, both tree diagram and formula came up with six combinations of 12 outcomes in which the order of the balls does not matter. Also refer to Diagram 1 and Example 6 of the Statistical Note 15 for detailed elaboration of this case.

Sampling With Replacement

A bag contains three balls of different colors (A of Blue color, B Green and C Red). Select two balls one randomly one at a time and replace back to the bag after selecting. Calculate the total number of outcomes containing two balls selected with replacement in which the order of balls does not matter.

Tree Diagram: This example has six combinations of nine outcomes each constituting two of all three balls, Blue (A), Green (B) and Red (C) selected one at random with replacement in which the order of balls does not matter. Each combination is indicated by the ball letters in a box and a number on the right most part of the Diagram 2.

Diagram 2: Outcomes in Selecting Two Out of Three Balls With Replacement in which the Order of Balls does not Matter

Formula: In above example, a formula (n+r-1)!/(n-1)!*r! is used (see Diagram  1 of Statistical Note 15). In this case, ‘n’ is equal to 3 and ‘r’ is equal to 2 balls. Using these values in the formula, we get 4!/2!*2!, which is equal to 6.

Thus, both tree diagram and formula came up with six combinations of nine outcomes consisting two balls in which the order of balls does not matter. Also refer to Example 8 of the Statistical Note 15 for detailed elaboration of this case.

The examples showed that the number of samples without or with replacement differed in which the position of the sampling units did not matter.