Past exam of the mathematics course of the University of Cambridge / 2020 / ia / Paper 2 / 1E / ii / Solution

There are $\lfloor n/2\rfloor$ even letters. The stated condition and injectivity force the even letters to permute among themselves, which can be done in $\lfloor n/2\rfloor !$ ways. The remaining odd letters must then permute among themselves. Thus the number is