
The number of words that can be written using all the letters of the word ‘IRRATIONAL’ is
$\eqalign{
& A)\dfrac{{10!}}{{{{(2!)}^3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,B)\dfrac{{10!}}{{{{(2!)}^2}}} \cr
& C)\dfrac{{10!}}{{(2!)}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,D)10! \cr} $
Answer
586.5k+ views
Hint:
If n be the total number of letters in a word and among which k number of letters are repeated by ${p_1}, {p_2}, {p_3},..., {p_k}$ times respectively,then the total number of distinct words using all the letters is
$\dfrac{{n!}}{{({p_1})!({p_2})!({p_3})!...({p_k})!}}$.
Complete step by step solution:
Step1: Here the given word is ‘IRRATIONAL’.
Step2: The total number of letters is $n = 10$.
Step3: The first letter ‘I’ occurs 2 times,then ${p_1} = 2$. The second letter ‘R’ occurs 2 times, then ${p_2} = 2$. The fourth letter ‘A’ occurs also 2 times, then ${p_3} = 2$.
Step4: All the other letters occur only once.
Step5: Then the number of words using all the letters will be
$\eqalign{
& \dfrac{{n!}}{{({p_1})!({p_2})!({p_3})!}} \cr
& = \dfrac{{10!}}{{(2)!(2)!(2)!}} \cr
& = \dfrac{{10!}}{{{{(2!)}^3}}} \cr} $
Hence, the option A) is correct here.
Note:
Let us verify through a small word ‘INN’. The total number of letters is $n = 3$. The second letter ‘N’ occurs 2 times, then ${p_1} = 2$. The other letter ‘I’ occurs only once. Then by the formula, the number of words using all the letters is $\dfrac{{n!}}{{({p_1})!}} = \dfrac{{3!}}{{2!}} = 3$. Hence three distinct words will be there. They are ‘INN’, ’NIN’ and ‘NNI’. These are all. Hence the formula holds.
If n be the total number of letters in a word and among which k number of letters are repeated by ${p_1}, {p_2}, {p_3},..., {p_k}$ times respectively,then the total number of distinct words using all the letters is
$\dfrac{{n!}}{{({p_1})!({p_2})!({p_3})!...({p_k})!}}$.
Complete step by step solution:
Step1: Here the given word is ‘IRRATIONAL’.
Step2: The total number of letters is $n = 10$.
Step3: The first letter ‘I’ occurs 2 times,then ${p_1} = 2$. The second letter ‘R’ occurs 2 times, then ${p_2} = 2$. The fourth letter ‘A’ occurs also 2 times, then ${p_3} = 2$.
Step4: All the other letters occur only once.
Step5: Then the number of words using all the letters will be
$\eqalign{
& \dfrac{{n!}}{{({p_1})!({p_2})!({p_3})!}} \cr
& = \dfrac{{10!}}{{(2)!(2)!(2)!}} \cr
& = \dfrac{{10!}}{{{{(2!)}^3}}} \cr} $
Hence, the option A) is correct here.
Note:
Let us verify through a small word ‘INN’. The total number of letters is $n = 3$. The second letter ‘N’ occurs 2 times, then ${p_1} = 2$. The other letter ‘I’ occurs only once. Then by the formula, the number of words using all the letters is $\dfrac{{n!}}{{({p_1})!}} = \dfrac{{3!}}{{2!}} = 3$. Hence three distinct words will be there. They are ‘INN’, ’NIN’ and ‘NNI’. These are all. Hence the formula holds.
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