How many 3 letter code words can be selected if repetition is not allowed?

There are four distinct letters in the word SERIES. You can either use three of the four letters or use two of the four letters by using either S or E twice.

You can use three distinct letters in $P(4, 3) = 4 \cdot 3 \cdot 2 = 24$ ways.

You can use exactly two letters if you use S or E twice. Thus, there are $C(2, 1)$ ways of choosing the repeated letter, $C(3, 2)$ ways of choosing where to place those letters in the three letter word, and $C(3, 1)$ of choosing the third letter in the word, giving

$$\binom{2}{1}\binom{3}{2}\binom{3}{1} = 2 \cdot 3 \cdot 3 = 18$$

ways to form a word with a repeated letter.

Consequently, there are $24 + 18 = 42$ distinguishable three letter words that can be formed with the letters of the word SERIES.

If a every outcome can be uniquely described via a sequence of answers to questions with number of choices $a_1, a_2, a_3,\dots$ respectively where regardless of which choice is made the number of available choices doesn't change, then the total number of outcomes will be $a_1\cdot a_2\cdot a_3\cdots$

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If letters are allowed to be repeated and order does matter:

Pick the first letter (5 choices), pick the second letter (5 choices), pick the third letter (5 choices), for a total of $5^3$ different codes.

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If letters are not allowed to be repeated and order does matter:

Pick the first letter (5 choices), pick the second letter (4 remaining choices), pick the third letter (3 remaining choices), for a total of $5\cdot 4\cdot 3 = \frac{5!}{2!}$ different codes.

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If letters are allowed to be repeated and order doesn't matter:

Apply stars-and-bars to get that there will be $\binom{3+5-1}{5-1}$ different codes.

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If letters are not allowed to be repeated and order doesn't matter:

Standard combinations problem: there are $\binom{5}{3}$ such choices. (Provable via a multiplication principle + division by amount of symmetry argument).

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Answer : (i) 60 (ii) 125

Solution : (i) Total number of 3-letter words is equal to the number of ways of filling 3 places. First place can be filled in 5 ways by any of the given five letters. Second place can be filled in 4 ways by any of the remaining 4 letters and the third place can be filled in 3 ways by any of the remaining 3 letters.
So, the required number of 3-letter words `=(5xx4xx3)=60.`
(ii) When repetition of letters is allowed, each place can be filled by any of the 5 letters in 5 ways.
`therefore " the required number of ways "=(5xx5xx5)=125.`

Hint: Here the given question is based on the concept of permutation. Here we have to find how many 3 letter codes can be formed using the first 10 letters of English alphabet and the repetition of letters is not allowed. We use permutation concepts and we determine the solution for the question.

Complete step-by-step solution:
In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set.
Now consider the given question,
 Here we have the first 10 letters of English alphabet, and need to find the number of ways where the 3 letters code can be formed without repeating any letters.
There are 10 letters of English alphabet.
As we know that repetition of the letters is not allowed,
so, the first place can be filled by any of 10 letters.
 Second place can be filled with any of the remaining 9 letters.
The third place can be filled with any of the remaining 8 letters.
Therefore, Number of 3-letter code can be formed when the repetition of letters is not allowed
\[ \Rightarrow \,\,\,10 \times 9 \times 8\]
On multiplication, we get
\[\therefore \,\,\,720\,\,ways\]
Therefore a total of 720 ways is possible.

Note: We can also find these type of question by directly using the permutation formula i.e., \[^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}\], Here, “\[^n{P_r}\]” represents the “r” objects to be selected from “n” objects without repetition, in which the order matters.
In the given question we have taken 3 letters to be selected from 10 letters without repetition, then by formula
\[ \Rightarrow \,\,{\,^{10}}{P_3} = \dfrac{{10!}}{{\left( {10 - 3} \right)!}}\]
\[ \Rightarrow \dfrac{{10!}}{{7!}}\]
\[ \Rightarrow \dfrac{{10 \times 9 \times 8 \times 7!}}{{7!}}\]
On simplification, we get
\[\therefore \,\,\,720\,ways\]

How many 3

How many 3

How many 3 alphabet arrangements are there with at least one repetition of arrangements )?