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ra1l [238]
3 years ago
6

there are 6 times as many ways to arrange the letters in the word "sleeplessness" as there are to arrange the letters in the wor

d "senselessness". Prove that this statement is true without actually calculating the number of ways the letters in each of the two words can be arranged.
Mathematics
2 answers:
EastWind [94]3 years ago
8 0

Using arrangements of words, it is found that the statement is true.

------------------

  • Suppose a word has n letters.
  • Considering that m of these letters are repeating, each n_1,n_2,...,n_m times.
  • The number of distinct ways the word can be arranged is given by:

N = \frac{n!}{n_1\times n_2 \times ... n_m}

------------------

  • The word "sleeplessness" has 13 letters.
  • s repeats 5 times.
  • e repeats 4 times.
  • l repeats 2 times.

Thus, the number of ways to arrange the letters is given by:

N_1 = \frac{13!}{5!4!2!}

------------------

  • The word "senselessness" has 13 letters.
  • s repeats 6 times.
  • e repeats 4 times.
  • n repeats 2 times.

Thus, the number of ways to arrange the letters is given by:

N_2 = \frac{13!}{6!4!2!}

------------------

Finding the ratio of N_2 to N_1:

\frac{N_2}{N_1} = \frac{\frac{13!}{5!4!2!}}{\frac{13!}{6!4!2!}} = \frac{13!}{5!4!2!} \times \frac{6!4!2!}{13!} = \frac{6!}{5!} = 6

Thus, since the ratio is 6, the statement is true.

A similar problem is given at brainly.com/question/16790460

rodikova [14]3 years ago
8 0

Answer:

This statement is true

Step-by-step explanation:

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