♦ Derivation of the formula for `text()^nP_r.`

♦ Permutations when all the objects are not distinct objects

♦ Permutations when all the objects are not distinct objects

`=>color(red)(text()^nP_r=(n!)/((n-r)!) , \ \ \ \ \ \ 0 ≤ r ≤ n)`

Let us now go back to the stage where we had determined the following formula:

`color(color)(text()^nP_r= n (n – 1) (n – 2) . . . (n – r + 1))`

Multiplying numerator and denomirator by `(n – r) (n – r – 1) . . . 3 × 2 × 1, ` we get

`text()^nP_r= (n (n – 1) (n – 2) . . . (n – r + 1) (n – r) (n – r – 1) . . . 3 × 2 × 1 )/ ((n – r) (n – r – 1) . . . 3 × 2 × 1, )=(n!)/((n-r)!)`

Thus `color(blue)(text()^nP_r=(n!)/((n-r)!))`` \ \ \ \ \ \ `where `0 ≤ r ≤ n`

This is a much more convenient expression for `text()^nP_r` than the previous one.

In particular, when `r = n,`

`\color{green} ✍️ color(red)(text()^nP_n=(n!)/(0!)=n!)`

Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have

`text()^nP_0 =1 = (n!)/(n!) = ( n!) / ((n -0)!)` .......................(1)

Therefore, the formula (1) is applicable for r = 0 also.

`text()^nP_r = (n!)/((n -r ) !), 0 <= r <= n`

`color{blue} " Theorem 2"` The number of permutations of n different objects taken r at a time, where repetition is allowed, is `n^r`.

Proof is very similar to that of Theorem 1 and is left for the reader to arrive at.

Here, we are solving some of the problems of the pervious Section using the formula for `text()^nP_r` to illustrate its usefulness.

In Example 1, the required number of words `= text()^4P_4 = 4! = 24`. Here repetition is not allowed. If repetition is allowed, the required number of words would be 44 = 256.

The number of 3-letter words which can be formed by the letters of the word NUMBER ` = text()^6 P_3 = (6!)/(3!) = 4 × 5 × 6 = 120`. Here, in this case also, the repetition is not allowed. If the repetition is allowed,the required number of words would be `6^3 = 216`

The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly `text()^12 P_2 = (12!)/(10!) = 11 xx 12 =132`

Let us now go back to the stage where we had determined the following formula:

`color(color)(text()^nP_r= n (n – 1) (n – 2) . . . (n – r + 1))`

Multiplying numerator and denomirator by `(n – r) (n – r – 1) . . . 3 × 2 × 1, ` we get

`text()^nP_r= (n (n – 1) (n – 2) . . . (n – r + 1) (n – r) (n – r – 1) . . . 3 × 2 × 1 )/ ((n – r) (n – r – 1) . . . 3 × 2 × 1, )=(n!)/((n-r)!)`

Thus `color(blue)(text()^nP_r=(n!)/((n-r)!))`` \ \ \ \ \ \ `where `0 ≤ r ≤ n`

This is a much more convenient expression for `text()^nP_r` than the previous one.

In particular, when `r = n,`

`\color{green} ✍️ color(red)(text()^nP_n=(n!)/(0!)=n!)`

Counting permutations is merely counting the number of ways in which some or all objects at a time are rearranged. Arranging no object at all is the same as leaving behind all the objects and we know that there is only one way of doing so. Thus, we can have

`text()^nP_0 =1 = (n!)/(n!) = ( n!) / ((n -0)!)` .......................(1)

Therefore, the formula (1) is applicable for r = 0 also.

`text()^nP_r = (n!)/((n -r ) !), 0 <= r <= n`

`color{blue} " Theorem 2"` The number of permutations of n different objects taken r at a time, where repetition is allowed, is `n^r`.

Proof is very similar to that of Theorem 1 and is left for the reader to arrive at.

Here, we are solving some of the problems of the pervious Section using the formula for `text()^nP_r` to illustrate its usefulness.

In Example 1, the required number of words `= text()^4P_4 = 4! = 24`. Here repetition is not allowed. If repetition is allowed, the required number of words would be 44 = 256.

The number of 3-letter words which can be formed by the letters of the word NUMBER ` = text()^6 P_3 = (6!)/(3!) = 4 × 5 × 6 = 120`. Here, in this case also, the repetition is not allowed. If the repetition is allowed,the required number of words would be `6^3 = 216`

The number of ways in which a Chairman and a Vice-Chairman can be chosen from amongst a group of 12 persons assuming that one person can not hold more than one position, clearly `text()^12 P_2 = (12!)/(10!) = 11 xx 12 =132`

Suppose we have to find the number of ways of rearranging the letters of the word ROOT. In this case, the letters of the word are not all different. There are 2 Os, which are of the same kind. Let us treat, temporarily, the 2 Os as different, say, `O_1` and `O_2`. The number of permutations of 4-different letters, in this case, taken all at a time is 4!. Consider one of these permutations say, `RO_1O_2T`. Corresponding to this permutation,we have 2 ! permutations `RO_1O_2T` and `RO_2O_1T` which will be exactly the same permutation if `O_1` and `O_2` are not treated as different, i.e., if `O_1` and `O_2` are the same O at both places.

Therefore, the required number of permutations `= (4!)/(2!) = 3 xx 4 = 12`

Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.

Temporarily, let us treat these letters different and name them as `I_1, I_2, T_1 , T_2, T_3`. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, `I_1 NT_1 SI_2 T_2 U E T_3`. Here if `I_1, I_2` are not same and `T_1, T_2, T_3` are not same, then `I_1, I_2` can be arranged in 2! ways and `T_1, T_2, T_3` can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation `I_1NT_1SI_2T_2UET_3`. Hence, total number of different permutations will be `(9!)/(2 ! 3!)`

We can state (without proof) the following theorems:

`color(red)("Theorem 3")` The number of permutations of n objects, where p objects are of the same kind and rest are all different `color(red)(=(n!)/(p !).)`

In fact, we have a more general theorem

`color(red)("Theorem 4")` The number of permutations of `color(blue)(n)` objects, where `color(blue)(p_1)` objects are of one kind, `color(blue)(p_2)` are of second kind, ........., `color(blue)(p_k)` are of `k^(th)` kind and the rest, if any, are of different kind is `color(red)((n!)/ (p_1! p_2! ...p_k!))`

Therefore, the required number of permutations `= (4!)/(2!) = 3 xx 4 = 12`

Let us now find the number of ways of rearranging the letters of the word INSTITUTE. In this case there are 9 letters, in which I appears 2 times and T appears 3 times.

Temporarily, let us treat these letters different and name them as `I_1, I_2, T_1 , T_2, T_3`. The number of permutations of 9 different letters, in this case, taken all at a time is 9 !. Consider one such permutation, say, `I_1 NT_1 SI_2 T_2 U E T_3`. Here if `I_1, I_2` are not same and `T_1, T_2, T_3` are not same, then `I_1, I_2` can be arranged in 2! ways and `T_1, T_2, T_3` can be arranged in 3! ways. Therefore, 2! × 3! permutations will be just the same permutation corresponding to this chosen permutation `I_1NT_1SI_2T_2UET_3`. Hence, total number of different permutations will be `(9!)/(2 ! 3!)`

We can state (without proof) the following theorems:

`color(red)("Theorem 3")` The number of permutations of n objects, where p objects are of the same kind and rest are all different `color(red)(=(n!)/(p !).)`

In fact, we have a more general theorem

`color(red)("Theorem 4")` The number of permutations of `color(blue)(n)` objects, where `color(blue)(p_1)` objects are of one kind, `color(blue)(p_2)` are of second kind, ........., `color(blue)(p_k)` are of `k^(th)` kind and the rest, if any, are of different kind is `color(red)((n!)/ (p_1! p_2! ...p_k!))`

Q 3078734606

Find the number of permutations of the letters of the word ALLAHABAD.

Here, there are 9 objects (letters) of which there are 4A’s, 2 L’s and rest are all different.

Therefore, the required number of arrangements `(9!)/(4! 2!) = (9xx8xx7xx6xx5)/2 = 7560`

Q 3048834703

How many 4-digit numbers can be formed by using the digits 1 to 9 if repetition of digits is not allowed?

Here order matters for example 1234 and 1324 are two different numbers. Therefore, there will be as many 4 digit numbers as there are permutations of 9 different digits taken 4 at a time.

Therefore, the required 4 digit numbers `text()^9P_4 = (9!)/{(9-4)!} = 9xx8xx7xx6 = 3024`

Q 3008034808

How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed?

Every number between 100 and 1000 is a 3-digit number. We, first, have to count the permutations of 6 digits taken 3 at a time. This number would be `text()^6P_3`. But, these permutations will include those also where `0` is at the `100’s` place. For example, 092, 042, . . ., etc are such numbers which are actually 2-digit numbers and hence the number of such numbers has to be subtracted from `text()^6P_3` to get the required number. To get the number of such numbers, we fix 0 at the 100’s place and rearrange the remaining 5 digits taking 2 at a time. This number is `text()^5P_2`. So

The required number ` = text()^6P_3 - text()^5P_2 = (6!)/(3!) - (5!)/(3!)`

` = 4xx5xx6-4xx5 = 100`

Q 3058134904

Find the value of n such that

(1). `text()^nP_5 = 42 text()^nP_3 , n > 4`

(2). `(text()^nP_4)/(text()^(n-1)P_4) = 5/3 , n > 4`

(1). `text()^nP_5 = 42 text()^nP_3 , n > 4`

(2). `(text()^nP_4)/(text()^(n-1)P_4) = 5/3 , n > 4`

(i) Given that `text()^nP_5 = 42 text()^nP_3 `

or `n (n – 1) (n – 2) (n – 3) (n – 4) = 42 n(n – 1) (n – 2)`

Since `n > 4` so `n (n-1) (n-2) ne 0`

Therefore, by dividing both sides by `n(n – 1) (n – 2),` we get

`(n – 3 (n – 4) = 42`

or `n^2 – 7n – 30 = 0`

or `n^2 – 10n + 3n – 30`

or `(n – 10) (n + 3) = 0`

or `n – 10 = 0` or `n + 3 = 0`

or `n = 10` or `n = – 3`

As n cannot be negative, so `n = 10.`

(ii). Given that `(text()^nP_4)/(text()^(n-1)P_4) = 5/3`

Therefore `3n (n – 1) (n – 2) (n – 3) = 5(n – 1) (n – 2) (n – 3) (n – 4)`

or `3n = 5 (n – 4) [as (n – 1) (n – 2) (n – 3) ≠ 0, n > 4]`

or `n = 10.`

Q 3078145006

Find `r`, if `5 text()^4P_r = 6 text()^5P_(r–1) .`

We have `5 text()^4P_r = 6 text()^5P_(r−1)`

or `5 xx (4!)/{(4-r)!} = 6xx(5!)/((5-r+1)!)`

or `(5!)/((4-r)!) = (6xx5!)/((5-r+1)(5-r)(5-r-1)!)`

or `(6 – r) (5 – r) = 6`

or `r^2 – 11r + 24 = 0`

or `r^2 – 8r – 3r + 24 = 0`

or `(r – 8) (r – 3) = 0`

or `r = 8 or r = 3`.

Hence `r = 8, 3.`

Q 3048245103

Find the number of different 8-letter arrangements that can be made from the letters of the word DAUGHTER so that

(i) all vowels occur together (ii) all vowels do not occur together.

(i) all vowels occur together (ii) all vowels do not occur together.

(i) There are 8 different letters in the word DAUGHTER, in which there are 3 vowels, namely, A, U and E. Since the vowels have to occur together, we can for the time being, assume them as a single object (AUE). This single object together with 5 remaining letters (objects) will be counted as 6 objects. Then we count permutations of these 6 objects taken all at a time. This number would be 6P6 = 6!. Corresponding to each of these permutations, we shall have 3! permutations of the three vowels A, U, E taken all at a time . Hence, by the multiplication principle the required number of permutations =` 6 ! × 3 ! = 4320.`

we first have to find all possible arrangments of 8 letters taken all at a time, which can be done in 8! ways. Then, we have to subtract from this number, the number of permutations in which the vowels are always together. Therefore, the required number `8 ! – 6 ! × 3 ! = 6 ! (7×8 – 6) = 2 × 6 ! (28 – 3) = 50 × 6 ! = 50 × 720 = 36000`

Q 3078245106

In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colour are indistinguishable ?

Total number of discs are `4 + 3 + 2 = 9.` Out of 9 discs, 4 are of the first kind

(red), 3 are of the second kind (yellow) and 2 are of the third kind (green). Therefore, the number of arrangements `(9!)/(4!3!2!) = 1260`

Q 3028345201

Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements,

(i) do the words start with P

(ii) do all the vowels always occur together

(iii) do the vowels never occur together

(iv) do the words begin with I and end in P?

(i) do the words start with P

(ii) do all the vowels always occur together

(iii) do the vowels never occur together

(iv) do the words begin with I and end in P?

There are 12 letters, of which N appears 3 times, E appears 4 times and D appears 2 times and the rest are all different. Therefore

The required number of arrangements `= (12 !)/(3! 4! 2!) = 1663200`

(i) Let us fix P at the extreme left position, we, then, count the arrangements of the remaining 11 letters. Therefore, the required of words starting with P are

`(11 !)/(3! 2! 4!) = 138600`

(ii) There are 5 vowels in the given word, which are 4 Es and 1 I. Since, they have to always occur together, we treat them as a single object `square` for the time being. This single object together with 7 remaining objects will account for 8 objects. These 8 objects, in which there are 3Ns and 2 Ds, can be rearranged in `(8!)/(3! 2!)` ways. Corresponding to each of these arrangements, the 5 vowels E, E, E, E and I can be rearranged in `(5!)/(4!)` ways. Therefore, by multiplication principle the required number of arrangements

` = (8!)/(3! 2!) xx (5!)/(4!) = 16800`

(iii) The required number of arrangements

= the total number of arrangements (without any restriction) – the number of arrangements where all the vowels occur together.

`= 1663200 – 16800 = 1646400`

(iv) Let us fix I and P at the extreme ends (I at the left end and P at the right end).

We are left with 10 letters.

Hence, the required number of arrangements

`(10!)/(3! 2! 4!) = 12600`