Question

How do you simplify \[\dfrac{{\left( {n - 1} \right)!}}{{n!}}\] ?

Answer 1

Hint: Factorial is product of numbers in increasing order. \[n\] is the last number whereas \[n - 1\] is the last but one number. Here the ratio of these two is asked to solve. So we can either directly solve this or can take any example for the value of \[n\]. Then on solving we will get to know the answer.

Complete step-by-step solution:
Given the ratio is \[\dfrac{{\left( {n - 1} \right)!}}{{n!}}\]
Now we can write \[n! = 1 \times 2 \times 3.... \times n\]
And \[\left( {n - 1} \right)! = 1 \times 2 \times 3.... \times n - 1\]
But \[n!\] can also be written as \[n! = 1 \times 2 \times 3.... \times n - 1 \times n\]
Now taking the ratio of both the numbers we get,
\[\dfrac{{\left( {n - 1} \right)!}}{{n!}} = \dfrac{{1 \times 2 \times 3.... \times n - 1}}{{1 \times 2 \times 3.... \times n - 1 \times n}}\]
Now we can observe that the terms in numerator and denominators are almost the same except \[n\] .
So we can write,
\[\dfrac{{\left( {n - 1} \right)!}}{{n!}} = \dfrac{1}{n}\]

Therefore the answer for this question is \[\dfrac{{\left( {n - 1} \right)!}}{{n!}} = \dfrac{1}{n}\]

Note: Another method:
Let \[n\] is 3.
Then we can write the ratio as, \[\dfrac{{\left( {3 - 1} \right)!}}{{3!}}\]
On simplifying it we get,
\[\dfrac{{2!}}{{3!}}\]
Thus we will write,
\[\dfrac{{1 \times 2}}{{1 \times 2 \times 3}}\]
Now the ratio becomes,
\[\dfrac{1}{3}\]
 That is \[\dfrac{1}{n}\]. This is the same answer.

Generally factorials are used in permutation and combinations. In these permutations the arrangement is done that can be of sitting of people, letters in a word, arrangement of vowels and consonants etc. and in combinations there is selection of objects, cards in a pack, balls from a basket etc. is done.