Aptitude Discussion

Q. |
In how many ways can the letters of the word $MANAGEMENT$ be rearranged so that the two $A$s do not appear together? |

✖ A. |
10! - 2! |

✖ B. |
9! - 2! |

✖ C. |
10! - 9! |

✔ D. |
$\text{None of these}$ |

**Solution:**

Option(**D**) is correct

The word $MANAGEMENT$ is a 10 letter word.

Normally, any 10 letter word can be rearranged in $10!$ ways.

However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters $A, M, E$ and $N$ repeat twice each.

Therefore, the number of ways in which the letters of the word $MANAGEMENT$ can be rearranged reduces to: $\dfrac{10!}{2! × 2! × 2! × 2!}$.

The problem requires us to find out the number of outcomes in which the two $A$s do not appear together.

The number of outcomes in which the two $A$s appear together can be found out by considering the two $A$s as one single letter.

Therefore, there will now be only 9 letters of which three of them $E, N$ and $M$ repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in: $\dfrac{9!}{2! × 2! × 2!}$ ways.

Therefore, the required answer in which the two $A$s do not appear next to each other,

$=\text{Total number of outcomes} - \text{the number of outcomes in which the 2 A’s appear together}$

$⇒ \dfrac{10!}{2! × 2! × 2! × 2!} -\dfrac{9!}{2! × 2! × 2!} \text{ways.}$