Question

the password for a website must have 8 characters consisting of letters and numbers with no repeated characters. what is the probability that a randomly selected password will have only numbers? enter your answer in the box.

Explanation:

Step1: Determine total possible characters

There are 10 digits (0 - 9) and 26 letters (A - Z, assuming case - insensitive or we can consider upper - and lower - case, but usually for such problems, if not specified, we can consider 26 letters). So total number of characters available is \(10 + 26=36\) (if we consider case - sensitive, it would be \(10+52 = 62\), but since the problem doesn't specify, we'll go with 26 letters for simplicity as a common approach in basic combinatorics problems of this type).

We need to form a password of 8 non - repeated characters. The total number of ways to choose 8 non - repeated characters from 36 (or 62) and arrange them (since the order of characters in a password matters) is given by the permutation formula \(P(n,r)=\frac{n!}{(n - r)!}\), where \(n\) is the total number of items, \(r\) is the number of items to be chosen and arranged.

For the total number of possible 8 - character passwords with no repeated characters (using \(n = 36\) and \(r=8\)): \(P(36,8)=\frac{36!}{(36 - 8)!}=\frac{36!}{28!}=36\times35\times34\times33\times32\times31\times30\times29\)

Step2: Determine number of favorable outcomes (passwords with only numbers)

There are 10 digits (0 - 9). The number of ways to form an 8 - character password with no repeated numbers is given by the permutation formula with \(n = 10\) and \(r = 8\) (since we need 8 non - repeated numbers from 10). So \(P(10,8)=\frac{10!}{(10 - 8)!}=\frac{10!}{2!}=10\times9\times8\times7\times6\times5\times4\times3\)

Step3: Calculate the probability

The probability \(P\) that a randomly selected password has only numbers is the number of favorable outcomes divided by the total number of possible outcomes.

\(P=\frac{P(10,8)}{P(36,8)}=\frac{\frac{10!}{2!}}{\frac{36!}{28!}}=\frac{10!×28!}{2!×36!}\)

Simplify the factorials:

\(36! = 36\times35\times34\times33\times32\times31\times30\times29\times28!\)

So \(\frac{10!×28!}{2!×36\times35\times34\times33\times32\times31\times30\times29\times28!}=\frac{10!}{2!×36\times35\times34\times33\times32\times31\times30\times29}\)

\(10! = 10\times9\times8\times7\times6\times5\times4\times3\times2\times1\), and \(2! = 2\times1\)

\(\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{2\times1\times36\times35\times34\times33\times32\times31\times30\times29}\)

Simplify numerator and denominator:

\(10\times9\times8\times7\times6\times5\times4\times3\times2\times1=10! = 3628800\)

\(2\times1\times36\times35\times34\times33\times32\times31\times30\times29=2\times36\times35\times34\times33\times32\times31\times30\times29\)

\(36 = 4\times9\), \(35 = 5\times7\), \(34 = 2\times17\), \(33 = 3\times11\), \(32 = 2^5\), \(30 = 3\times10\), \(29\) is prime.

But let's calculate the values:

\(P(10,8)=10\times9\times8\times7\times6\times5\times4\times3=\ 1814400\)

\(P(36,8)=36\times35\times34\times33\times32\times31\times30\times29\)

\(36\times35 = 1260\), \(1260\times34 = 42840\), \(42840\times33 = 1413720\), \(1413720\times32 = 45239040\), \(45239040\times31 = 1402410240\), \(1402410240\times30 = 42072307200\), \(42072307200\times29 = 1220096908800\)

\(P(10,8) = 1814400\)

So the probability \(P=\frac{1814400}{1220096908800}=\frac{1814400\div1814400}{1220096908800\div1814400}=\frac{1}{6724520}\) (approximate value, we can also simplify the fraction step - by - step:

\(\frac{10\times9\times8\times7\times6\times5\times4\times3}{36\times35\times34\times33\times32\times31\times30\times29}=\frac{(10\times9\times8\times7\times6\times5\times4\times3)}{(36\times35\times34\times33\times32\times31\times30\times29)}\)

\(36 = 4\times9\), \(35 = 5\times7\), \(30 = 3\times10\)

Cancel out common factors:

9 in numerator and denominator: \(\frac{10\times\cancel{9}\times8\times7\times6\times5\times4\times3}{(4\times\cancel{9})\times(5\times7)\times34\times33\times32\times31\times(3\times10)\times29}\)

7 in numerator and denominator: \(\frac{10\times8\times\cancel{7}\times6\times5\times4\times3}{4\times(5\times\cancel{7})\times34\times33\times32\times31\times3\times10\times29}\)

5 in numerator and denominator: \(\frac{10\times8\times6\times\cancel{5}\times4\times3}{4\times\cancel{5}\times34\times33\times32\times31\times3\times10\times29}\)

10 in numerator and denominator: \(\frac{\cancel{10}\times8\times6\times4\times3}{4\times34\times33\times32\times31\times3\times\cancel{10}\times29}\)

4 in numerator and denominator: \(\frac{8\times6\times\cancel{4}\times3}{\cancel{4}\times34\times33\times32\times31\times3\times29}\)

3 in numerator and denominator: \(\frac{8\times6\times\cancel{3}}{\ 34\times33\times32\times31\times\cancel{3}\times29}\)

Now we have \(\frac{8\times6}{34\times33\times32\times31\times29}\)

\(8\times6 = 48\)

\(34\times33 = 1122\), \(1122\times32 = 35904\), \(35904\times31 = 1113024\), \(1113024\times29 = 32277696\)

\(\frac{48}{32277696}=\frac{1}{672452}\) (Wait, I think I made a mistake in the earlier cancellation. Let's do it again:

Original fraction: \(\frac{10P8}{36P8}=\frac{\frac{10!}{2!}}{\frac{36!}{28!}}=\frac{10!×28!}{2!×36!}=\frac{10!}{2!×36×35×34×33×32×31×30×29}\)

\(10! = 3628800\), \(2! = 2\), \(36×35 = 1260\), \(34×33 = 1122\), \(32×31 = 992\), \(30×29 = 870\)

\(36×35×34×33×32×31×30×29=1260×1122×992×870\)

\(1260×1122 = 1413720\), \(1413720×992 = 1402410240\), \(1402410240×870 = 1220096908800\)

\(2!×36×35×34×33×32×31×30×29=2×1220096908800 = 2440193817600\)

\(\frac{3628800}{2440193817600}=\frac{3628800\div3628800}{2440193817600\div3628800}=\frac{1}{6724520}\) (Ah, I see, earlier when I canceled 36 as 4×9, I missed that 36 is also 6×6, but the correct way is to use the permutation values.

\(P(10,8)=\frac{10!}{(10 - 8)!}=\frac{10!}{2!}=10\times9\times8\times7\times6\times5\times4\times3 = 1814400\)

\(P(36,8)=\frac{36!}{(36 - 8)!}=\frac{36!}{28!}=36\times35\times34\times33\times32\times31\times30\times29 = 36\times35 = 1260\), \(1260\times34 = 42840\), \(42840\times33 = 1413720\), \(1413720\times32 = 45239040\), \(45239040\times31 = 1402410240\), \(1402410240\times30 = 42072307200\), \(42072307200\times29 = 1220096908800\)

So \(\frac{1814400}{1220096908800}=\frac{1814400\div1814400}{1220096908800\div1814400}=\frac{1}{6724520}\)

Answer:

\(\frac{1}{6724520}\) (or approximately \(1.487\times10^{-7}\))