Permutations and Combinations
3.0 Permutation
3.0 Permutation
Statement: The number of permutations of '$n$' distinct objects, taken '$r$' at a time, is the total number of arrangements of '$n$' distinct objects, in groups of '$r$' where the order of arrangement is important.
'Permutation in other words is the arrangement of objects in an order.'
There are basically two conditions:
- Without repetition: Here the objects are unique.
- With repetition: Here the objects need not be unique.
$\left( i \right)$ Without repetition:
(A) Arrangement of $n$ distinct objects, taking $r$ at a time.
| $n$ | $n-1$ | $n-2$ | ....... | $n-(r-1)$ |
Number of ways in which it can be arranged is same as number of ways of filling $r$ places
$$\begin{equation} \begin{aligned} = n(n - 1)(n - 2)(n - 3).....(n - r + 2)(n - r + 1) \\ \\ = \frac{{n(n - 1)(n - 2).....(n - r + 2)(n - r + 1)(n - r)!}}{{\left( {n - r} \right)!}} \\ \\ = \frac{{n!}}{{\left( {n - r} \right)!}} \\ \\ { = ^n}{P_r} \\\end{aligned} \end{equation} $$
(B) Arranging $n$ objects, taking all at a time, $${ = ^n}{P_n} = n!$$
$\left( ii \right)$ With repetition:
(A) Number of permutations of $n$ objects taken $r$ at a time, where each object can repeat itself upto $r$ number of times is = ${n^r}$
| $n$ | $n$ | $n$ | ....... | $n$ |
(B) Number of permutations possible for $n$ objects out of which $p$ are identical and of one kind and $q$ are identical and of one kind and the rest are unique is, $$={{n!} \over {p!q!}}$$
Proof: Let there be $n$ objects to be arranged, where $p$ are identical and $q$ are identical.
Total number of permutations considering each one to be distinct is $n!$,
$$\begin{equation} \begin{aligned} {\{ {a_1},{a_2},.......,{a_p}\} _p}{\{ {b_1},{b_2},.......,{b_q}\} _q}{c_1}{c_2}{c_3}.....{c_{n - p - q}} \\ {\{ {a_2},{a_1},.......,{a_p}\} _p}{\{ {b_3},{b_2},.......,{b_q}\} _q}{c_1}{c_2}{c_3}.....{c_{n - p - q}} \\ ............. \\ {\{ {a_5},{a_6},.......,{a_p}\} _p}{\{ {b_4},{b_7},.......,{b_q}\} _q}{c_1}{c_2}{c_3}.....{c_{n - p - q}} \\\end{aligned} \end{equation} $$
In this arrangement '$p!q!$' ways are one and the same.
Let there be $y$ unique arrangements possible. Thus total number of ways is, $$(p!)(q!)y$$
$$n!=y(p!)(q!)$$
Number of unique permutations possible is given by, $$y={{n!} \over {p!q!}}$$
Question 7. How many $9$ letter, $5$ letter and $4$ letter words with or without meaning can be formed using the word EDUCATION?
Solution: The word EDUCATION has $9$ unique letters. Hence words are formed without repetition.
$9$ letter words: $=9!=3,62,880$ words.
$5$ letter words: $=^9{P_5}$= ${9!} \over {(9-5)!}$ = ${{9!} \over {4!}}=15,120$ words.
$4$ letter words: ${ = ^9}{P_4} = \frac{{9!}}{{(9 - 4)!}} = \frac{{9!}}{{5!}} = 3024$ words.
Question 8. Consider the word 'THINKMERIT'.
$\left( i \right)$ How many anagrams can be made using the letters of the word THINKMERIT?
$\left( ii \right)$ Out of these, how many are such that all the vowels always come together?
$\left( iii \right)$ How many of these of these start and end with a vowel?
Solution: $\left( i \right)$ There are $10$ letters in the word THINKMERIT.
Out of these 'T' is repeated twice and 'I' is repeated twice.
Thus the total arrangements possible is given as,
$$= \frac{{10!}}{{2!2!}} = 9,07,200$$
$\left( ii \right)$ There are three vowels.
We can write as THNKMRT (IIE).
Considering them to be one unit, we have $8$ units in which 'T' is repeated twice.
Hence, ${{8!} \over {2!}}$ ways.
The vowels can be arranged in ${{3!} \over {2!}}$.
Hence total ways is, $$=\frac{{8!}}{{2!}} \times \frac{{3!}}{{2!}} = 60,480$$
$\left( iii \right)$ The words can be, E...I, I...I, I...E , Thus $3$ ways.
There are ${{8!} \over {2!}}$ ways for other letters including one vowel.
Thus total words possible, $$ = \frac{{8!}}{{2!}} \times 3 = 60,480$$
Question 9. If all the letters of the word $ARRANGE$ are arranged in all possible ways, how many of these will have the $A$'s not together and also $R$'s not together?
Solution: There are $7$ letters, out of which $A$ and $R$ are repeated twice.
Thus the total possible words $$={{7!} \over {2!2!}}$$
Let us consider the two $A$'s always occur together. Thus the word can be seen as $(AA)RRNGE$. Now there are $6$ letters and $R$ is repeated twice. Thus the total possible words are, $$={{6!} \over {2!}}$$
Let us consider the two $R$'s always occur together. Thus the word can be seen as $(RR)AANGE$. Now there are $6$ letters and $A$ is repeated twice. Thus the total possible words are, $$={{6!} \over {2!}}$$
Let us consider the two $A$'s always occur together and the two $R$'s always occur together. Thus the word can be seen as $(AA)(RR)NGE$. Now there are $5$ letters. Thus the total possible words are, $={5!}$.
The total number of arrangements possible for the condition = total possible words - {total possible word when $A$'s occur together + total possible word when $R$'s occur together - total possible ways in which $A$'s occur together and $R$'s occur together}
$$\begin{equation} \begin{aligned} = \frac{{7!}}{{2!2!}} - \left[ {\frac{{6!}}{{2!}} + \frac{{6!}}{{2!}} - 5!} \right] \\ = 1260 - \left[ {720 - 120} \right] \\ = 660 \\\end{aligned} \end{equation} $$
Question 10. If all the letters of the following words are arranged in all possible manner as in a dictionary, then find their rank in the same.
a. $QUEUE$
b. $INDIA$
Solution:
a. QUEUE
- Arranging it alphabetical order, EEUUQ.
- Number of words beginning with:
1. [E] EUUQ $={{4!} \over {2!}}=12$
2. [QE] EUU $ = \frac{{3!}}{{2!}} = 3$
3. [QU] EEU = QUEEU
- And the next word being QUEUE.
- Thus there are, $12 + 3 + 1 = 16$ words preceding QUEUE.
- Thus the rank of QUEUE is ${17}^{th}$.
b. INDIA
- Arranging it alphabetical order, ADIIN.
- Number of words beginning with:
1. [A] DIIN $ = \frac{{4!}}{{2!}} = 12$
2. [D] AIIN $={{4!} \over {2!}}=12$
- The set of words begin with [I] i.e. the same letter as that of the original word, thus considering the succeeding letter
- Number of words beginning with:
1. [IA] DIN $ = 3! = 6$
2. [ID] AIN $ = 3! = 6$
3. [II] ADN $ = 3! = 6$
- Now the set of words begin with [IN] i.e. the same letters as that of the original word, thus considering the succeeding letter
1. [INA] DI $=2!$ = $2$
- Now the set of words begin with [IND] i.e. the same letters as that of the original word, thus considering the succeeding letter
1. [INDA] I , there is $1$ way.
- The next word being INDIA.
- Thus there are, $12 + 12 + 6 + 6 + 6 + 2 + 1 = 45$ words preceding INDIA.
- Thus the rank of INDIA is ${46}^{th}$.
