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The letters of the word DECIMAL are randomly arranged into a new 'word' also consisting of seven letters. How many different arrangements are possible if:

The arrangements must start with a vowel and end in a consonant and repetition of letters is allowed.
in Mathematics by Diamond (88,180 points) | 145 views

1 Answer

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Answer:

There are 3 vowels $\Rightarrow 3$ options for first position

There are 4 consonants $\Rightarrow 4$ options for last position.

The remaining 5 letters can be arranged in $5 \times 4 \times 3 \times 2 \times 1$ ways

$3 \times(5 \times 4 \times 3 \times 2 \times 1) \times 4=1440$

Explanation:

This problem involves higher order reasoning as there is no obvious route to the solution. You need to analyse the given in formation and conceptually understand what you must do in order to distil a strategy to solve this problem. In this problem you need to distinguish the vowels from the consonants in the given word and then establish that there are three vowels and 4 consonants. You need to reason that as there are three vowels, there are then 3 options for the first position. Similarly, as there are 4 consonants there are then 4 options of the last position. More importantly and complex, is that you must come to see that there are five remaining positions and that these positions can be filled in by the five remaining letters in 5 ways (i.e. 5x $4 \times 4 \times 2 \times 1$ ways) and this requires conceptual understanding of the arrangement of the new 'word', which consists of seven letters abiding to the given conditions. Finally, within the context of the counting principle you need to put all the sub-aspects and elements together, to finally establish that the number of different arrangements can be expressed as follows: $3 \times(5 \times 4 \times 3 \times 2 \times 1) \times 4=1440$. 

by Diamond (88,180 points)

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