Question

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

Answer 1

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

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• 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:

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

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• 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:

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

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• 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:

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

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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

Answer 2

Answer:

This statement is true

Step-by-step explanation: