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Consider the four letters A, B and C. Find the number of ordered arrangement when considered 2 letters at a time if (i) replacement is allowed (ii) replacement is not allowed

Mathematics

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Answers

2 parts
i

9\text{9}

ii

6\text{6}

Step-by-step solution

4 steps

The given letters are A, B, and C. So, the total number of distinct letters is n=3n=3.
We need to find the number of ordered arrangements when considering 2 letters at a time, so r=2r=2.

Part i· 2 steps

Replacement is allowed

  1. Determine the number of choices for each position.

    Since replacement is allowed, for the first letter, there are 3 choices (A, B, or C). For the second letter, there are also 3 choices (A, B, or C), as the first letter can be repeated.

  2. Calculate the total number of ordered arrangements.

    The number of ordered arrangements with replacement is given by nrn^r.

    32=93^2 = 9

    The possible arrangements are: AA, AB, AC, BA, BB, BC, CA, CB, CC.

    The number of ordered arrangements when replacement is allowed is 9\boxed{\text{9}}.

Part ii

6\text{6}

Replacement is not allowed

  1. Determine the number of choices for each position.

    Since replacement is not allowed, for the first letter, there are 3 choices (A, B, or C). For the second letter, there are only 2 remaining choices, as the first letter cannot be repeated.

  2. Calculate the total number of ordered arrangements.

    The number of ordered arrangements without replacement is given by the permutation formula P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n-r)!}.

    P(3,2)=3!(3−2)!=3!1!=3×2×11=6P(3, 2) = \frac{3!}{(3-2)!} = \frac{3!}{1!} = \frac{3 \times 2 \times 1}{1} = 6

    The possible arrangements are: AB, AC, BA, BC, CA, CB.

    The number of ordered arrangements when replacement is not allowed is 6\boxed{\text{6}}.

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