Let's blame Eve

We use a technique called stars and bars , which determines the number of ways to put $k$ indistinguishable balls in $n$ distinguishable boxes. Here is a summary. Draw $n-1$ bars. I will draw 7-1=6.

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Now, say all objects to the left of the first bar are type A, those to the left of bar 2 are type B, and so on, until those to the right of the last divider, which are type G. Therefore, this corresponds to our task. We distribute $k$ stars (here, 4).

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The number of ways to do this is $\dbinom{n-1+k}{k}$ . This is the formula.

Now, to the problem. Fix one fruit to be from the Tree of Life. Eve now has 3 fruits left to pick, and 7 trees to pick from. The number of ways to do this, by stars and bars, is $\dbinom{7+3-1}{3}=\dbinom{9}{3}=\boxed{84}$

Mathematically, the question is equivalent to find number of solution to this equation-

$Tree_{1} + Tree_{2} + Tree_{3} + Tree_{4} + Tree_{5} + Tree_{6} + Tree_{7} = 3$

where $Tree_{n}$ represent no of fruits taken from $n^{th}$ tree.

The number of solutions to this equation is equivalent to find the number of ways in which $3$ can be written down using $6$ $+$ signs.

For example one such way to write 3 is $1+1+1++++$ which would mean that three fruits are taken from first three trees and $0$ fruits are taken from other trees. Similarly, each new way to write $3$ would imply a different solution to the above equation.

Hence, the number of ways to do that are $\frac{9!}{6! 3!}$ $=$ $84$

There are 7 trees, and Eve must pick 3 fruits. Let $t_i$ equal the number of fruits eve picks from tree $i$ . Thus, $t_1 + t_2 + t_3 + t_4 + t_5 + t_6 + t_7 = 3$ . This is equivalent to ${{7 + 3 - 1} \choose 3} = 84$ .

Note: The formula for giving $n$ things to $k$ distinct people is ${{n + k - 1} \choose n}$

We can ignore the Tree of Knowledge of good and evil. Let $L$ represent the Tree of Life and let the $6$ other trees be represented by a number between $1$ and $6$ . We can consider 4 cases for this problem:

Case 1 - $4$ apples are picked from $L$ :

In this case, we can see that there is only $1$ combination (an apple is chosen from $L$ each time).

Case 2 - $3$ apples are picked from $L$ :

In this case, we need the number of ways $6$ can fill $1$ space. This is just ${6 \choose 1} = 6$ combinations.

Case 3- $2$ apples are picked from $L$ :

We need to consider $2$ situations in this case: if $1$ or $2$ apples are chosen from each tree. If $2$ apples are chosen from a tree, then there are $6$ combinations. If $1$ apple is chosen from each tree, then there are ${6 \choose 2} = 15$ combinations. So the total number of combinations for case $3$ is $15 + 6 = 21$ .

Case 4- $1$ apple is picked from $L$ :

We need to consider $3$ situations in this case: if $1$ , $2$ or $3$ apples are chosen from each tree. If $3$ apples are chosen from a tree, then there are $6$ combinations. If $2$ apples are chosen from a tree, then there are $6 \times 5 = 30$ combinations, and when $1$ apple is chosen from each other tree, then there are ${6 \choose 3} = 20$ combinations. So the total number of combinations for case $4$ is $6 + 30 + 20 = 56$ .

Therefore, the total number of combinations is $1 + 6 + 21 + 56 = \fbox{84}$

Let the Abiu tree be A, the Babaco tree be B, the Cherimoya tree be C, the Durian tree be D, the Emblic tree be E, the Feijoa tree be F, the Tree of Knowledge of good and evil be K, and the Tree of Life be L. We know that there are no K's, and there is at least one L. We will do casework based on the number of L's.

Case 1 (4 L's (Such as LLLL)): There is only $\boxed{1}$ way to do this, namely all L's.

Case 2 (3 L's and another fruit (Such as LLLD)): There are $\boxed{6}$ ways to do this, because there are 6 ways to pick the other fruit.

Case 3 (2 L's):

Case 3a: (2 L's and 2 of the same other fruit (Such as LLDD)): There are $\boxed{6}$ ways to do this, since there are 6 ways to pick the other fruit.

Case 3b: (2 L's and 2 different other fruits (Such as LLDF)): There are $\boxed{15}$ ways to do this, because you can pick the 2 other fruits in $\dbinom{6}{2}=15$ different ways.

Case 4: (1 L)

Case 4a: (1 L and 3 of the same other fruit (Such as LDDD)): There are $\boxed{6}$ ways to do this, because you can pick the other fruit in 6 ways.

Case 4b: (1 L, 1 of a different fruit, and 2 of another fruit different from the first 2 (Such as LDDF)): There are $\dbinom{6}{2}=15$ ways to pick the 2 different fruits, and 2 ways to pick which one we want 2 of. (For example, LDDF is different from LDFF), for a total of $\boxed{30}$ ways.

Case 4c: (One L and 3 other different fruits (Such as LDEF)): There are $\boxed{20}$ ways to do this, namely the $\dbinom{6}{3}=20$ ways to pick the 3 other fruits.

Since there must be at least one L, all cases are covered, and we have a total of $1+6+6+15+6+30+20=\boxed{\boxed{84}}$ different ways to choose 4 fruits.

Lemma The number of ways of choosing $m$ elements from a set of $n$ distinct elements allowing repetition but ignoring permutations is ${m+n-1 \choose m}$ .

Proof Let the elements of the set be labelled $\{a_1, a_2, \cdots , a_n \}$ . For each $i$ , suppose $b_i$ instances of the element $a_i$ appear in the set of $m$ selections. Then we obtain the following equation: $b_1 + b_2 + \cdots + b_n= m$ Here each $b_i$ s are non-negative integers. We now have a bijection: each solution to the above equation corresponds to a unique selection of $m$ elements. But by the ball and bars method, we know that this equation has ${m+n-1 \choose m}$ solutions. QED

We now go back to our original problem. Note that the Tree of Knowledge of Good and Evil is irrelevant to the problem: we might as well assume that this tree does not exist, and proceed hereupon. We divide the problem into three cases.

Case $1$ : Eve chooses exactly $1$ fruit from the Tree of Life
Then Eve has to choose $3$ fruits from $6$ remaining trees, allowing repetition but ignoring order. By our previous lemma, this can be done in ${6+3-1 \choose 3}$ ways.

Case $2$ : Eve chooses exactly $2$ fruits from the Tree of Life
Now she has to choose $2$ fruits from $6$ remaining trees. Again by the lemma, she can do this in ${6+2-1 \choose 2}= 21$ ways.

Case $3$ : Eve chooses exactly $3$ fruits from the Tree of Life
Now there is $1$ fruit remaining, and she has to choose it from $6$ trees. This can be obviously done in $6$ ways.

Case $4$ : Eve chooses exactly $4$ fruits from the Tree of Life
The selection is then uniquely determined: this can be done in $1$ way.

We then add the possibilities to get the final result: $56+21+6+1= \boxed{84}$ .

We use a technique called stars and bars, which determines the number of ways to put indistinguishable balls in distinguishable boxes. Here is a summary. Draw bars. I will draw 7-1=6. | | | | | | Now, say all objects to the left of the first bar are type A, those to the left of bar 2 are type B, and so on, until those to the right of the last divider, which are type G. Therefore, this corresponds to our task. We distribute stars (here, 4). | * | * | | | * | * The number of ways to do this is . This is the formula. Now, to the problem. Fix one fruit to be from the Tree of Life. Eve now has 3 fruits left to pick, and 7 trees to pick from. The number of ways to do this, by stars and bars, is

Since one fruit is always to be picked from the tree of Life (L) and no fruit is to be picked from the three of Knowledge (K), the problem reduces to picking $3$ fruits from $7$ trees viz. L, A, B, C, D, E, F.

Let's arrange the fruits picked by Eve according to their type in the order L, A, B, C, D, E, F. Let us transform fruits of different colors in a plain-fruit (p) and in order to distinguish the type associated with this plain-fruit, lets put a bar between the fruits of different types.

Hence, if Eve picked EEF, we represent it using plain-fruits notation as:

| | | | | p p |p

Since any collection of fruits from Eve can be translated into this representation, the total ways in which Eve can pick up $3$ fruits from $7$ trees $=$

Total ways to choose $3$ 'p' (or $6$ '|') out of $9$ slots $= 9C_{3} = 84$

Since there are 7 trees to choose from (excluding the Tree of Knowledge of good and evil) there are:

(7+6+5+4+3+2+1)+(6+5+4+3+2+1)+(5+4+3+2+1)+(4+3+2+1)+(3+2+1)+(2+1)+(1)=84

Here's a breadth first search approach:

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# starting fruit count is "start"
# note we've already picked a
# fruit from the tree of life and
# we're avoiding the Tree of KoGaE
start = (1, 0, 0, 0, 0, 0, 0)
frontier = [start]

# pick 3 fruits
for i in range(3):
    new = set()
    # ...for each known possibility
    for state in frontier:
        # ...from 7 trees
        for n in range(7):
            adj = list(state)
            adj[n] += 1
            new.add(tuple(adj))
    frontier = new
print ("Answer:", len(frontier))