What is the probability of the random arrangement of letters in the word university and two?

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Concept:

Arrangement of n things in which r things are likely is given as \(\frac{{{\rm{n}}!}}{{{\rm{r}}!}}\)

Calculation:

We have 10 letters word “UNIVERSITY”. So, the arrangement of 10 things can be done in 10! ways. But, two I’s are repeating. So, the arrangement can be done in \(\frac{{10!}}{{2!}} = 5 \times 9!\)

Number of ways of arrangement that the two I’s do not come together = Total number of arrangement – number of arrangement in which both I’s are together.

Number of arrangements in which both I’s are together:

We consider both I’s as one letter. Therefore, we now have 9 letters. So the number of arrangement = 9!

∴ Number of ways of arrangement that the two I’s do not come together = 5 × 9! – 9!

= 4 × 9!

So, the required probability \( = \frac{{{\rm{number\;of\;arrangements\;two\;I'snot\;together}}}}{{{\rm{total\;number\;of\;arrangemets}}}}\)

\(= \frac{{4 \times 9!}}{{5 \times 9!}}\)

= 4/5

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Find the probability that in a random arrangement of the letters of the word 'UNIVERSITY', the two I's do not come together.

Out of the letters in the word ‘UNIVERSITY’, there are two I’s.
Number of permutations = \[\frac{10!}{2}\]

The number of words in which two I’s are never together is given by
total number of words – number of words in which two I’s are together.

\[= \frac{10!}{2} - 9!\]\[ = \frac{10! - 2 \times 9!}{2}\]\[ = \frac{9!\left[ 10 - 2 \right]}{2}\]\[ = \frac{9! \times 8}{2}\]

\[ = 9! \times 4\]