Question

suppose we want to choose 2 letters, without replacement, from the 5 letters A, B, C, D, and E (a) How many ways can this be done, if the order of the choices is relevant? (b) How many ways can this be done, if the order of the choices is not relevant?

Answer 1

To select 2 letters from 5 with order relevance, there are 20 ways using permutations. Without order relevance, there are 10 ways using combinations. These mathematical concepts help distinguish the total number of different outcomes for a given scenario.

Step-by-step explanation:

Combinations and Permutations

To answer the question, we need to understand the concepts of combinations and permutations in mathematics. The difference between the two is whether or not the order matters.

(a) Permutations (Order Matters)

To choose 2 letters from 5 (A, B, C, D, E) with order being relevant, we are dealing with permutations. This can be calculated using the permutation formula, which is nPr = n! / (n - r)! where 'n' is the total number of options and 'r' is the number of options being chosen. In this case, n=5 and r=2.

The calculation for permutations is 5P2 = 5! / (5 - 2)! = 5 × 4 = 20 ways.

(b) Combinations (Order Does Not Matter)

To choose 2 letters from the 5 without considering the order, we are dealing with combinations. This can be calculated using the combination formula, which is nCr = n! / [(n - r)! × r!]. Therefore, 5C2 = 5! / [(5 - 2)! × 2!] = 5! / (3! × 2!) = (5 × 4) / (2 × 1) = 10 ways.

Answer 2

Answer: a. 20 and b.10

Step-by-step explanation:

Given Letters = A, B, C, D, and E

Number of letters given = 5

a. When we choose 2 letters, without replacement from the 5 letters and order of the choices is relevant we use permutations.

Then, the number of ways we can choose 2 letters from 5 =
`^5P_2`

`=(5!)/((5-2)!)=(5*4*3!)/(3!)=20` [∵
`^nP_r=(n!)/((n-r)!)`]

So , when order of the choices is relevant , the total number of way to do this = 20

b. When we choose 2 letters, without replacement from the 5 letters and order of the choices is not relevant we use combinations.

Then, the number of ways we can choose 2 letters from 5 =
`^5C_2`

`=(5!)/(2!(5-2)!)=(5*4*3!)/(2*3!)=10` [∵
`^nC_r=(n!)/(r!(n-r)!)`]

So , when order of the choices is not relevant , the total number of way to do this = 10