In how many ways can be the letter of the word ‘STRANGE’ be arranged so that
[a] The vowel may appear in the odd places
[b] The vowels are never separated
[c] The vowels never come together
Answer
Hint:There are 7 letters in the word ‘STRANGE’ in which two are vowels. There are four odd places. So, we have to find the total number of ways for these four places. The other 5 letters may be placed in the five places in 5! ways. By this we get the total number of required arrangements. If vowels are never separated, in this case consider all the vowels a single letter. If all the vowels never come together in this case, we subtract all vowels together from the total number of
arrangements.Complete step-by-step answer:
For finding number of ways of placing n things in r objects, we use ${}^{n}{{P}_{r}}=\dfrac{n!}{\left[ n-r \right]!}$. Also, for arranging n objects without restrictions, we have $n!$ ways.
[a] There are 7 letters in the word ‘STRANGE’ amongst which 2 are vowels and there are 4 odd places [1, 3, 5, 7] where these two vowels are to be placed together.
Total number of ways can be expressed by replacing n = 4 and r = 2,
$\begin{align}
&
{}^{4}{{P}_{2}}=\dfrac{4!}{\left[ 4-2 \right]!} \\
& =\dfrac{4!}{2!} \\
& =\dfrac{4\times 3\times 2!}{2!} \\
& =4\times 3=12 \\
\end{align}$
And corresponding to these 12 ways the other 5 letters may be placed in 5! ways $=5\times 4\times 3\times 2\times 1=120$.
Therefore, the total number of required arrangements = $12\times 120=1440$ ways.
[b] Now, vowels are not to be separated. So, we consider all vowels as a single letter.
There are
six letters S, T, R, N, G, [AE] , so they can arrange themselves in 6! ways and two vowels can arrange themselves in 2! ways.
Total number of required arrangements \[=6!\times 2!=6\times 5\times 4\times 3\times 2\times 1\times 2\times 1=1440\] ways.
[c] Now, for the number of arrangements when all vowels never come together, we subtract the total arrangement where all vowels have occurred together from the total number of arrangements.
The total number of arrangements = 7! = 5040
ways
And the number of arrangements in which the vowels do not come together $=7!-6!2!$
number of arrangements in which the vowels do not come together $=5040 -1440 = 3600$ ways.
Note: Knowledge of permutation and arrangement of objects under some restriction and without restriction is must for solving this problem. Students must be careful while making the possible cases. They must consider all the permutations in word like in part [b], a common mistake is done by not considering the
arrangements of vowels.
In how many different ways can the letters of the word TRAINER be arranged so that the vowels always come together?
A. 1440
B. 120
C. 720
D. 360
Answer
Verified
Hint: To solve this problem we have to know about the concept of permutations and combinations. But here a simple concept is used. In any given word, the number of ways we can arrange the word by jumbling the letters is the number of letters present in the word factorial. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$ \Rightarrow n! = n[n - 1][n - 2].......1$
Complete step by step answer:
Given the word
TRAINER, we have to arrange the letters of the word in such a way that all the vowels in the word TRAINER should be together.
The number of vowels in the word TRAINER are = 3 vowels.
The three vowels in the word TRAINER are A, I, and E.
Now these three vowels should always be together and these vowels can be in any order, but they should be together.
Here the three vowels AIE can be arranged in 3 factorial ways, as there are 3 vowels, as given below:
The number of ways the 3
vowels AIE can be arranged is = $3!$
Now arranging the consonants other than the vowels is given by:
As the left out letters in the word TRAINER are TRNR.
The total no. of consonants left out are = 4 consonants.
Now these 4 consonants can be arranged in the following way:
As in the 4 letters TRNR, the letter R is repeated for 2 times, hence the letters TRNR can be arranged in :
$ \Rightarrow \dfrac{{4!}}{{2!}}$
But the letters TRNR are arranged along with the vowels A,I,E,
which should be together always but in any order.
Hence we consider the three vowels as a single letter, now TRNR along with AIE can be arranged in:
$ \Rightarrow \dfrac{{5!}}{{2!}}$
But here the vowels can be arranged in $3!$ as already discussed before.
Thus the word TRAINER can be arranged so that the vowels always come together are given below:
$ \Rightarrow \dfrac{{5!}}{{2!}} \times 3! = \dfrac{{120 \times 6}}{2}$
$ \Rightarrow 360$
The number of ways the word TRAINER can be arranged so that the vowels always come together are 360.
Note: Here while solving such kind of problems if there is any word of $n$ letters and a letter is repeating for $r$ times in it, then it can be arranged in $\dfrac{{n!}}{{r!}}$ number of ways. If there are many letters repeating for a distinct number of times, such as a word of $n$ letters and ${r_1}$ repeated items, ${r_2}$ repeated items,…….${r_k}$ repeated items, then it is arranged in $\dfrac{{n!}}{{{r_1}!{r_2}!......{r_k}!}}$ number of ways.