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1.How many natural-number factors does n have if N = 2^4 * 3^3 * 5^2 * 7^2?

 

2.How many distinct sequences of four letters can be made from the letters in EQUALS if each sequence must begin with L, end with Q, and no letter can appear in a sequence more than once?

Guest Mar 11, 2018
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2+0 Answers

 #1
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 1)  N = 2^4 * 3^3 * 5^2 * 7^2?

Don't quite understand your question! If you multiply them together, you get 529,200 = 2^4 * 3^3 * 5^2 * 7^2. The number of factors is simply the sum of the exponents that you already have: 4+3+2+2=11 factors, but only 4 DISTINCT factors: 2, 3, 5, 7.

Guest Mar 11, 2018
edited by Guest  Mar 11, 2018
 #2
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2)

Here is my attempt at this:

E Q U A L S

Since the first and the last letters are fixed at L and Q respectively, then the remaining 2 letters out of 4 can be permuted in 4P2 =12 ways as follows:

LAEQ, LASQ, LAUQ, LEAQ, LESQ, LEUQ, LSAQ, LSEQ, LSUQ, LUAQ, LUEQ, LUSQ (12)

Guest Mar 11, 2018

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