Question

the following questions involve arrangements of letters.
number of ways to arrange a, b, c, d, e in a line =
number of ways to arrange a, b, c, d, e in a circle =
number of ways to arrange a, b, c, d, d in a line =
number of ways to arrange 3 out of the 5 letters (a, b, c, d, e) in a line =
number of ways to arrange 3 out of the 5 letters (a, b, c, d, e) in a circle =

Explanation:

Step1: Line permutation of 5 distinct letters

This is a full permutation of 5 unique elements. The formula for permutations of $n$ distinct items is $n!$.
$5! = 5\times4\times3\times2\times1 = 120$

Step2: Circular permutation of 5 distinct letters

For circular permutations of $n$ distinct items, the formula is $(n-1)!$ (since rotations are considered identical).
$(5-1)! = 4! = 4\times3\times2\times1 = 24$

Step3: Line permutation with repeated letters

We have 5 total letters where 2 are identical (D). The formula is $\frac{n!}{k!}$, where $n=5$ and $k=2$ (number of repeated items).
$\frac{5!}{2!} = \frac{5\times4\times3\times2\times1}{2\times1} = 60$

Step4: Line permutation of 3 out of 5 distinct letters

This is a partial permutation, calculated by $P(n,r)=\frac{n!}{(n-r)!}$, where $n=5$, $r=3$.
$P(5,3)=\frac{5!}{(5-3)!} = \frac{5\times4\times3\times2\times1}{2\times1} = 60$

Step5: Circular permutation of 3 out of 5 distinct letters

First choose 3 letters from 5, then arrange them in a circle. The formula is $\binom{n}{r}\times(r-1)!$, which simplifies to $\frac{n!}{r\times(n-r)!}$.
$\frac{5!}{3\times(5-3)!} = \frac{5\times4\times3\times2\times1}{3\times2\times1} = 20$

Answer:

120
24
60
60
20