A meeting of the minds!

There are two possibilities for the order of $A$ through $D$ :

• $A > B > C > D$
• $A > B > D > C$

This gives $10$ equiprobable possibilities for the order of $A$ through $E$ , since $E$ can fit in anywhere in either of the two scenarios above. (Five possible positions in each)

Only $3$ result in $D>E$ . Namely $E$ being youngest in the first one, or $E$ being second or first youngest in the second one.

Therefore $P(D>E) = \dfrac{3}{10}$

$3+10 = \boxed{13}$

(The fact that there are $20$ people is irrelevant)

10 possibilities: 1)ABCDE 2)ABDCE 3)ABCED 4)ABDEC 5)ABECD 6)ABEDC 7)AEBCD 8)AEBDC 9)EABCD 10)EABDC and only 1),2) and 4), is D older than E, therefore having a probability of 3/10

Using the given information, there are two possibilities: $A > B > C > D$ Or $A > B > D > C$ Let (1) denote the first case, and (2) the second. Assuming (1) does happen, we observe that adding Edward gives us five different possibilities. Because we need to save the relation between the four people. Though only 1 gives us the scenario where D > E. Meaning the probability is $\frac{1}{5}$ . Doing the same for the second case, we get the probability of $\frac{2}{5}$ . So the total probability is: $\frac{1}{2} * \frac{1}{5} + \frac{1}{2} * \frac{2}{5} = \frac{3}{10}$

Only 3 (abcde, abdce, abdec) out of the 5! possible combinations of the 5 people satisfy the requirements, out of the 2 (abcd, abdc) out of 4! already given. Thus, 3/120 ÷ 2/24 = 3/10

Since this is probability, we can ignore all other people and focus on the five mentioned in the question because each permutation of them appears in an equal number of permutation of all the people.

So before we include Edward, if we organize them in order of age there are two possibilities: $ABCD$ and $ABCD$ .

$E$ has five possible locations to be added in each of these, providing a total of $10$ possibilities. The only three where David is older are:

$ABCDE$ , $ABDEC$ and $ABDCE$ .

So the probability of David being older will be $\frac {3}{10}$ . $3$ and $10$ are coprime, so:

$3+10=\boxed {13}$

The total number is irrelevant; we only care that there are five named people. The probability that someone other than Edward (call this person Allen) is oldest is $\frac{4}{5}$ . The probability that someone other than Edward is next oldest (call them Bob) is $\frac{3}{4}$ . Carrie's age relative to Edward's is unknown and therefore irrelevant, but there is a $\frac{1}{2}$ probability that Dave is older than Edward. $\frac{4}{5}\times\frac{3}{4}\times\frac{1}{2} = \frac{12}{40} = \frac{3}{10}$

( 3×(20!/5!) ) / ( 2×(20!/4!) ) =3/10

I guessed it. Stop reading this. (By the way, this is just a joke comment)

Brute force python:

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from itertools import permutations as perm

l = [] 
conf = perm(range(20),5)
for a,b,c,d,e in conf:
    if a > b and b > c and b > d:
        l.append((a,b,c,d,e))
m = len(l)  

n = 0
for _,_,_,d,e in l:                          
    if d > e:
        n += 1

a, b = n, m
while a:
    a, b = b % a, a
print(n//b + m//b)