Aptitude Discussion

Q. |
In how many rearrangements of the word $AMAZED$, is the letter $E$ positioned in between the 2 $A$s (Not necessarily flanked)? |

✖ A. |
$24$ |

✖ B. |
$72$ |

✔ C. |
$120$ |

✖ D. |
$240$ |

**Solution:**

Option(**C**) is correct

In any rearrangement of the word, consider only the positions of the letters $A, A$ and $E$.

These can be as $A A E$, $A E A$ or $E A A$.

So, effectively one-third of all words will have $E$ in between the two $A$s.

The total number of rearrangements are $\dfrac{6!}{2!} = 360$

One-third of $360$ is,

$=\dfrac{1}{3}\times360$

$=\textbf{120}$

**Mohit**

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**OoooO**

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let, A,A,E as one

so (AAE)MZD = 4 digits

now they can be rearranged in 4! ways

but AAE can also be rearranged among them in 3!/2! ways (as there's 2A)

so total ways= (3!/2!)x4! =72

And the answer is B.72

**Pritish**

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AEA can be arranged in 3!/2! ways and considering them as one AEA=x

XMZD these can be arranged in 4! ways therefore 4!*3!/2!= 36 should be total.

**Pradeep**

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But according to question E should be in between two A's.for that the arrangement will be like AEAMZD in this MZD can interchange their position and AEA together will change its position by this mean total no of ways should be $\dfrac{4!}{2!} =12$ ways 2! for repeated A.

You forgot the permutation for term related to AEA.

**Anirban**

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there is no gurantee that the letters $AAE$ will be together.

To make sure that that $AAE$ is together we need to consider them a bunch and then the total no of words will be 4!.

Now the above logic is applicable

Anirban it says 'Not necessarily flanked' so we can not consider them as a bunch.

Another method order is fixed (A M A) now we have to choose 3 out of 6 (6c3) and for rest three we can arrange it by 3!. so ans will be (6c3)* 3!