Calculation of number of permutations

Question

Find the number of permutations of word BANANA such that no two A's and N's are adjacent.

My attempt: Let

$A=$ No of permutations in which A is together.

$B=$ No of permutations in which N is together.

We want $A'\cap B'$ which is $U-(A\cup B).$

$A$ is $4!/2!$

$B$ is $5!/3!$

$A\cap B$ is $3!$

By PIE: $A\cup B = A+B-A\cap B = 26$

$U= 6!/3!2!$

Therefore the required answer is $60-26=34$. But the answer is given as $10.$ Where am I wrong?

Answer 1

Treat NA as one character, then there are $\frac{4!}{2!}=12$ ways of permuting B-A-NA-NA, including $2$ permutations containing NA-A that you want to omit, hence $10$.

Answer 2

You have three $A$ and you don't want them to be adjacent. On the other hand you have only letters N and B except that. Hence, at least you have one 'ANA'. This only indicates that you don't need to worry about the adjacent N anymore. Now you can break down the case:

• Starting with ANA: 4 cases ANA-(B,N)-[A-(N,B) or (N,B)-A]
• Ending with ANA: 4 cases similar to above
• in the middle: 2 cases: NANABA and ABANAN

Answer 3

You applied the Inclusion-Exclusion Principle incorrectly.

The number of arrangements of the letters of the word BANANA can be found by selecting three of the six positions for the A's, two of the remaining three positions for the N's, and then filling the final open position with the B, which can be done in $$\binom{6}{3}\binom{3}{2}\binom{1}{1} = \frac{6!}{3!3!} \cdot \frac{3!}{2!1!} \cdot \frac{1!}{1!0!} = \frac{6!}{3!2!1!}$$ ways, as you found.

From these, we must subtract those arrangements in which one or more pairs of adjacent letters are identical.

A pair of adjacent letters are identical: We consider cases, depending on whether a pair of A's or a pair of N's are adjacent.

A pair of A's are adjacent: We have five objects to arrange: AA, A, B, N, N. Choose two of the five positions for the N's. Arrange the remaining three distinct objects in the remaining three positions. There are $$\binom{5}{2}3!$$ such arrangements.

A pair of N's are adjacent: We have five objects to arrange: A, A, A, B, NN. Choose three of the five positions for the A's. Arrange the remaining two distinct objects in the remaining two positions. There are $$\binom{5}{3}2!$$ such arrangements.

Two pairs in which adjacent letters are identical: This can occur in two ways. Either the pairs of adjacent identical letters overlap, meaning the three A's are consecutive, or they are disjoint, in which case there is a pair of adjacent A's and a pair of adjacent N's.

Two overlapping pairs of adjacent identical letters: We have four objects to arrange: AAA, B, N, N. Choose two of the four positions for the N's. Arrange the remaining two distinct objects in the remaining two positions. There are $$\binom{4}{2}2!$$ such arrangements.

Two distinct pairs of adjacent identical letters: We have four objects to arrange: A, AA, B, NN. The four objects are all distinct, so they can be arranged in $4!$ ways.

Three pairs of adjacent identical letters: This can only occur if the three A's are consecutive and the two N's are consecutive. Thus, we have three objects to arrange: AAA, B, NN. Since they are distinct, they can be arranged in $3!$ ways.

By the Inclusion-Exclusion Principle, the number of admissible arrangements is $$\binom{6}{3}\binom{3}{2}\binom{1}{1} - \binom{5}{2}3! - \binom{5}{3}2! + \binom{4}{2}2! + 4! - 3!$$