10. The number of distinguishable arrangements of all the letters in the word BANANA, where the two N's cannot be together, is

Step 1 ：The word BANANA has 6 letters, with 3 A's, 2 N's and 1 B. The total number of arrangements of these letters without any restrictions is given by the formula for permutations of a multiset: \(\frac{6!}{3! * 2! * 1!} = 60\).

Step 2 ：We need to subtract the number of arrangements where the two N's are together. We can think of the two N's as a single entity, so we have 5 entities to arrange: {NN, A, A, A, B}. The number of arrangements of these entities is \(\frac{5!}{3! * 1! * 1!} = 20\).

Step 3 ：But within the entity {NN}, the two N's can be arranged in 2! = 2 ways. So the total number of arrangements where the two N's are together is 20 * 2 = 40.

Step 4 ：Therefore, the number of distinguishable arrangements of all the letters in the word BANANA, where the two N's cannot be together, is 60 - 40 = 20.

Step 5 ：Final Answer: The number of distinguishable arrangements of all the letters in the word BANANA, where the two N's cannot be together, is \(\boxed{20}\).