From all the permutations of 5 different letters from the 7 letters D, E, C, I, M, A, and L, how many begin and end with a consonant?

Let's first begin by finding out how many permutations there are with a population of 7 and we're choosing 5. The general formula for permutations is:

#P_(n,k)=(n!)/((n-k)!); n="population", k="picks"#

#(7!)/((7-5)!)=(7!)/(2!)=5040/2=2520#

Of these, how many begin and end with a consonant?

We can "glue" a consonant to the beginning and ending of the different words and thereby force them to be consonants. In how many ways can we do this? There are 2 positions we care about (first letter, last letter) and 4 consonants to choose from, so we can express that as:

#P_(4,2)=(4!)/((4-2)!)=(4!)/(2!)#

The remaining letters can now be arranged in whatever order, and so that's 5 letters remaining and we're picking 3:

#P_(5,3)=(5!)/((5-3)!)=(5!)/(2!)#

And so all together we have:

#(P_(5,3))(P_(4,2))=(5!)/(2!)(4!)/(2!)=((120)(24))/((2)(2))=120xx6=720#

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How many have A or C in the centre?

We can "pin" one of them in the centre of the word. There are 2 choices for that.

The rest of the word can be any of the 6 remaining letters (and we're picking 4 of them):

#2(P_(6,4))=((2)(6!))/((6-4)!)=((2)(6!))/(2!)=6! = 720#

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Alternating consonants and vowels

With this, we can either have CVCVC or VCVCV. We can calculate them and then add them together.

CVCVC - we have 4 consonants to pick from and we're choosing 3. We also have 3 vowels and are choosing 2:

#(P_(4,3))(P_(3,2))=((4!)(3!))/(((4-3)!)((3-2)!))=((4!)(3!))/((1!)(1!))=24xx6=144#

VCVCV - we have 4 consonants to pick from and we're choosing 2. We also have 3 vowels and are choosing all 3:

#(P_(4,2))(P_(3,3))=((4!)(3!))/(((4-2)!)((3-3)!))=((4!)(3!))/((2!)(0!))=(24xx6)/(2xx1)=72#

And so:

#144+72=216#