5 Marks Questions

Question 15 Marks

Find the number of words formed by permuting all the letters of the following words:
INDIA.

Answer

There are 5 letters in the word 'INDIA' out of which 2 are I'S, and the rest are all distinct.
so, the total number of
$=\frac{5!}{2!\ }$
$=\frac{5\times4\times3\times2!}{2!}$
$=60$

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Question 25 Marks

If P(n, 4) = 12 . P(n, 2), find n.

Answer

We have,
P(n, 4) = 12 . P(n, 2).
$\Rightarrow \frac{\text{n!}}{(\text{n-4})!}=12\times\frac{\text{n!}}{(\text{n-2)}! }\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow \frac{1}{(\text{n-4})!}=\frac{12}{(\text{n-2)}!}$
$\Rightarrow \frac{1}{(\text{n}-4)}!= \frac{12}{(\text{n}-2)(\text{n}-2 -1)(\text{n}-2-2)!}$
$\Rightarrow \frac{1}{(\text{n}-4)!}= \frac{12}{(\text{n}-2)(\text{n}-3)\text{n}-4)!}$
$\Rightarrow \frac{(\text{n}-2)(\text{n}-3)(\text{n}-4)!}{(\text{n}-4)!}= 12$
$\Rightarrow (\text{n}-2)(\text{n}-3) =12$
$\Rightarrow \text{n}^2-3\text{n}-2\text{n}-12=0$
$\Rightarrow \text{n}^2+5\text{n}-6=0$
$\Rightarrow \text{n}^2-6\text{n}+1\text{n}-6=0$
$\Rightarrow \text{n}(\text{n}-6)+1(\text{n}-6)= 0$
$\Rightarrow (\text{n}-6)(\text{n}+1)= 0$
$\Rightarrow \text{n}-6=0$
$\Rightarrow \text{n}=6 \ [\because \text{n}\neq-1]$
Hence, $\text{n}=6$

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Question 35 Marks

How many words $($with or without dictionary meaning$)$ can be made from the letters in the word $\text{MONDAY,}$ assuming that no letter is repeated, if

$4$ letters are used at a time?
All letters are used at a time.
All letters are used but first is vowel.

Answer

$\text{MONDAY}$ has $6$ letters with no repetitions, so

Number of words using $4$ letters at a time with no repetitions $=\ ^6\text{P}_4$
$=\frac{6!}{2!}$
$=360$

Number of words using all $6$ letters at a time with no repetitions $=\ ^6\text{P}_6$

$=\frac{6!}{(6-6)!}$
$= 6\times 5\times4\times3\times2\times1$
$=720$

Number of words using all $6$ letters, starting with vowels

$=2\times\ ^5\text{P}_5$
$= 2\times5\times4\times3\times2\times1$
$=240$

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Question 45 Marks

If P(11, r) = P(12, r − 1) find r.

Answer

We have,
P(11, r) = P(12, r − 1).
$\Rightarrow \frac{11!}{(11-\text{r})!}= \frac{12!}{\big[12-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow \frac{11!}{(11-\text{r})!}=\frac{12\times11!}{[12-\text{r}+1]!}$
$\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{[13-\text{r}]!}$
$\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{[13-\text{r}]!}$
$\Rightarrow \frac{1}{(11-\text{r})!}= \frac{12}{(13-\text{r})\times(13-\text{r}-1)(13-\text{r}-2)!}$
$\Rightarrow \frac{1}{(11-\text{r})!}=\frac{12}{(13-\text{r})\times(12-\text{r})(11-\text{r})!}$
$\Rightarrow \frac{(13-\text{r})(12-\text{r}){(11-\text{r})}}{(11-\text{r})}=12$
$\Rightarrow (13-\text{r})(12-\text{r})=12$
$\Rightarrow 156-13\text{r}-12\text{r}+\text{r}^2=12$
$\Rightarrow \text{r}^2-25\text{r}+156-12=0$
$\Rightarrow \text{r}^2-25\text{r}+144=0$
$\Rightarrow \text{r}^2-16\text{r}-9\text{r}+144=0$
$\Rightarrow \text{r}(\text{r}-16)-9(\text{r}-16)=0$
$\Rightarrow (\text{r}-9)(\text{r}-16)=0$
$\Rightarrow \text{r}-9=0\begin{bmatrix}\ \because\text{r}\ \leq\ 11 \\ \therefore \ \neq\ 16 \end{bmatrix}$
$\Rightarrow \text{r}=9$

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Question 55 Marks

Find the number of words formed by permuting all the letters of the following words:
CONSTANTINOPLE.

Answer

There are 14 letters in the word 'CONSTANTINOPLE' out of which 2 are O's, 3 are N's, 2 are T's and the rest are all distinct.
So, the total number of words
words $=\frac{14!}{2!\ 3!\ 2!}$
$=\frac{14!}{2\times 3\times2\times2}$
$=\frac{14!}{24}$

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Question 65 Marks

If the letters of the word 'LATE' be permuted and the words so formed be arranged as in a dictionary, find the rank of the word LATE.

Answer

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with A, E, L, Tin order.
'A' will occur in the first place as often as remaining 3 letters all at a time i .e A will occur in the first place the same number of times.
$\therefore$ Number of words starting with A = 3! = 6
Number of words starting with E = 3! = 6
Number of words begining with Lis 3!, but one of these words is the word LATE itself. The first word beginning with Lis the word LATE and the next word is LATE.
$\therefore$ Rank of LATE = 2 × 6 + 2
= 12 + 2
= 14

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Question 75 Marks

If P(9, r) = 3024, find r.

Answer

We have,
P(9, r) = 3024
${P}(9,\text{r}) = \frac{9!}{(9-\text{r})!}\Big[\because^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{3024}{9\times 8\times7\times6\times5\times4\times3\times2\times1}$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{336}{8\times7\times6\times5\times4\times3\times2\times1}$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{42}{7\times6\times5\times4\times3\times2\times1}$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{42}{7\times6\times5\times4\times3\times2\times1}$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{1}{5\times4\times3\times2\times1}$
$ \Rightarrow\frac{1}{(9-\text{r})!} =\frac{1}{5!}$
$\Rightarrow (9-\text{r})! =5!$
$\Rightarrow 9-\text{r}= 5$
$\Rightarrow 9-5=\text{r}$
$\Rightarrow 4=\text{r}$
$\Rightarrow \text{r}= 4$
Hence, $ \text{r}= 4 $

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Question 85 Marks

If P(2n − 1, n) : P(2n + 1, n − 1) = 22 : 7 find n.

Answer

We have,
P(2n − 1, n) : P(2n + 1, n − 1) = 22 : 7
$\Rightarrow \frac{\text{p}(\text{2n-1,n})}{\text{P} (\text{2n}+1,\text{n} -1)}=\frac{22}{7}$
$\Rightarrow\frac{\frac{\text{(2n-1)!}}{(\text{(2n-1-n)!})}}{\frac{\text{(2n+1)}!}{\text{[2n+1-(n-1)]!}}}=\frac{22}{7}$
$\Rightarrow \frac{\text{(2n-1) }\times\text{(n+2)!}}{\text{(n-1)!}(\text{2n+1})!}=\frac{22}{7}$
$\Rightarrow \frac{\text{(2n-1)}\times\text{(n+2)(n+2-1(n+2-2)(n+2-3)!}}{\text{(n+1)!(2n+1).2n.(2n+1)!}}=\frac{22}{7}$
$\Rightarrow \frac{\text{n(n+2)(n+1)}}{2\text {(2n+1)}}=\frac{22}{7}$
$\Rightarrow \frac{\text{n}^2+\text{n}+\text{2n}+2}{4\text{n}+2}=\frac{22}{7}$
$\Rightarrow 7(\text{n}^2+3\text{m}+2)=22\times (4\text{n}+2)$
$\Rightarrow7\text{n}^2+21\text{n}+44=88\text{n}-44$
$\Rightarrow7\text{n}^2+21\text{n}-88\text{n}+14-44=0$
$\Rightarrow 7\text{n}^2-67\text{n}-30=0$
$\Rightarrow 7\text{n}^2 -70\text{n}+3\text{n}-30=0$
$\Rightarrow 7\text{n}(\text{n}-10)+3(\text{n}-10)=0$
$\Rightarrow (\text{n}-10)(7\text{n}+3)=0$
$\Rightarrow \text{n}-10=0$
$\Rightarrow \text{n}=10\ [\because \text{7n+3}\neq 0]$

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Question 95 Marks

Find the number of words formed by permuting all the letters of the following words:
EXERCISES.

Answer

There are 9 letters in the word 'EXERCISES' out of which 3 are E's, 2 are S's and the rest are all distinct.
So, the total number of words
$=\frac{9!}{3!\ 2!}$
$=\frac{9\times8\times7\times6\times5\times4\times3!}{3!\times2\times1}$
$=9\times8\times7\times6\times5\times2$
$=30240$

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Question 105 Marks

Find the number of words formed by permuting all the letters of the following words:
INDEPENDENCE.

Answer

There are 12 letters in the word 'INDEPENDENCE' out of which 2 are D'S, 3 are N'S, 4 are E'S and the rest are all distinct.
so, the total number of words $=\frac{12!}{2!\ 3!\ 4!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4!}{2!\ 3!\ 4!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5}{2\times3\times2 }$
$=11\times10\times9\times8\times7\times6\times5$
$=1663200.$

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Question 115 Marks

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?

Answer

4 red, 3 yellow and 2 green discs.
Total discs = 9
Required number of ways
$=\frac{9!}{4!\ 3!\ 2!}$
$=\frac{9\times8\times7\times6\times5}{3\times2\times2}$
$=1260$
Required number of ways = 1260.

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Question 125 Marks

If a denotes the number of permutations of (x + 2) things taken all at a time, b the number of permutations of x things taken 11 at a time and c the number of permutations of x − 11 things taken all at a time such that a = 182 bc, find the value of x.

Answer

'a' denotes the number of permutations of (x + 2) things taken all at a time. $\therefore \ \text{a} = \ ^{\text{x}+2}\text{P}_{\text{x + 2}} $"b' is the number of permutations of x things taken 11 at a time.
$\therefore\text{b}= \ ^\text{x}\text{P}_{11}$
and, C is the number of permutations of x - 11 things taken all at a time.
$\therefore\ \text{c}= \ ^{ \text{x}-11}\text{P}_{\text{x}-11}$ Now, a = 182bc [Given] $\Rightarrow \ ^{\text{x}+2}\text{P}_{\text{x + 2}} =182\ \times\ ^\text{x}\text{P}_{11} \times\ ^{ \text{x}-11}\text{P}_{\text{x}-11}$ $\Rightarrow (\text{x+2})=182 \times \frac{\text{x!}}{(\text{x}-11)!}\times\text{(x-11)!}\\$ $ \bigg[ \because \ ^\text{n}\text{P}_\text{n}=\text{n!} \text{ and } ^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n-r}!)}\bigg]$ $\Rightarrow (\text{x}+2)! = 182\times\text{x}!$ $\Rightarrow (\text{x}+2)(\text{x}+1)\text{x}! = 182\times\text{x}!$ $\Rightarrow (\text{x}+2)(\text{x}+1) = 182$ $\Rightarrow \text{x}^2+\text{x}+2\text{x}+2 =182$ $\Rightarrow \text{x}^2+3\text{x}+2 -182=0$ $\Rightarrow \text{x}^2+3\text{x}-180=0$ $\Rightarrow \text{x}^2+15\text{x}-12\text{x}-180=0$ $\Rightarrow \text{x}(\text{x}+15)-12(\text{x}+15)=0$ $\Rightarrow (\text{x}-12)(\text{x}-15)=0$ $\Rightarrow \text{x}-12=0\ [\because \text{x}\neq-15]$ $\Rightarrow \text{x}=12$ Hence, $\text{x}=12$

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Question 135 Marks

Prove that:
1 . P(1, 1) + 2 . P(2, 2) + 3 . P(3, 3) + ... + n . P(n, n) = P(n + 1, n + 1) − 1.

Answer

We have,
1 . P(1, 1) + 2 . P(2, 2) + 3 . P(3, 3) + ... + n . P(n, n) = P(n + 1, n + 1) − 1
$=1.1+2.2!+3.3!.......\text{n}.\text{n}!\ [\because\text{p}(\text{n}.\text{n})=\text{n}!]$
$=\sum\limits_\text{r-1}^\text{n}\text{r}.\text{r}!$
$=\sum\limits_\text{r-1}^\text{n}\big[(\text{r+1})\text{r}!-\text{r}!\big]$
$=\sum\limits_\text{r-1}^\text{n}\big[(\text{r+1})!-\text{r}!\big]\ \big[\because(\text{r+1})\text{r}!-(\text{r+1})\text{r}!\big]$
$=\big[(2!-!)+(3!-2!)+(4!-3!)..........+(\text{n}+1)!-\text{n!}\big]$
$=(\text{n}+1)!-1!$
$=\ ^\text{n+1}\text{P}_\text{n-1}-1! \ [\because ^\text{n}\text{P}_\text{n}=\text{n}!]$
$\text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 145 Marks

A biologist studying the genetic code is interested to know the number of possible arrangements of 12 molecules in a chain. The chain contains 4 different molecules represented by the initials A (for Adenine), C (for Cytosine), G (for Guanine) and T (for Thymine) and 3 molecules of each kind. How many different such arrangements are possible?

Answer

Total number of molecules = 12
Now,
the chain contains 4 different molecules A, C, G, and T, and 3 molecules of each kind.
$\therefore$ The number of different arrangements $=\frac{12!}{3!\ 3!\ 3!\ 3!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3!}{3\times2\times3\times2\times3\times2\times3!}$
$=369600.$
Hence, the number of different possible arrangements are = 369600.

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Question 155 Marks

Find the number of words formed by permuting all the letters of the following words:
ARRANGE.

Answer

There are 7 letters in the word 'ARRANGE' out of which 2 are A'S, 2 are R'S, and the rest are all distinct.
so, the total number of words
$=\frac{7!}{2!\ 2!}$
$=\frac{7\times6\times5\times4\times3\times2!}{2!\ 2!}$
$=\frac{7\times6\times5\times4\times3}{2\times1 }$
$=7\times6\times5\times4\times3$
$=1260$

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Question 165 Marks

Find the number of words formed by permuting all the letters of the following words:
SERIES.

Answer

There are 6 letters in the word 'SERIES' out of which 2 are S's, 2 are E's and the rest are all distinct.
so, the total num bar of words
$=\frac{6!}{2!\ 2!}$
$=\frac{6\times5\times4\times3\times2!}{2!\ 2!}$
$=\frac{6\times5\times4\times3}{2\times1}$
$=6\times5\times2\times3$
$=180$

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Question 175 Marks

How many permutations can be formed by the letters of the word, $\text{'VOWELS',}$ when.

There is no restriction on letters?
Each word begins with $E?$
Each word begins with $O$ and ends with $L?$
All vowels come together$?$
All consonants come together$?$

Answer

There are $6$ letters in the word $\text{'VOWELS'.}$ The total number of words formed with these $6$ letters is the number of arrangements of $6$ items, taken all at a time, which is equal to $^6P_6 = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
If we fix up $E$ in the begining then the remaining $5$ letters can be arranged in $^5P_5 = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ways
If we fix up $O$ in the begining and Lat the end, the remaining $4$ letters can be arranged in $^4P_4 = 4! = 4 \times 3 \times 2 \times 1 = 24.$
There are $2$ vowels and $4$ consonants in the word $\text{'VOWELS'.}$

Considering $2$ vowels as one letter, we have letters which can be arranged in $^5p_5 = 5!$ ways. $O, E$ can be put together in $2!$ ways.
Hence, required number of words $= 5! \times 2!$
$= 5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1$
$= 120 \times 2$
$= 240$

There are $2$ vowels and $4$ consonants in the word $\text{'VOWELS'.}$

Considering $4$ consonants as one letter, we have $3$ letters which can be arranged in $^3P_3 = 3!$ ways.
$\text{U, W, L, S}$ can be put together in $4!$ ways.
Hence, required number of words in which all consonants come together $= 3! \times 4!$
$= 3 \times 2 \times 4 \times 3 \times 2$
$= 144$.

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Question 185 Marks

Find the total number of permutations of the letters of the word 'INSTITUTE'.

Answer

There are 9 letters in the word 'INSTITUTE' out of which 2 are I's, 3 are T's and the rest are all distinct.
$\therefore$ The total number of permutations of the letters of the word 'INSTITUTE' $=\frac{9!}{2!\ 3!}$
Hence, the total number of words are $\frac{9!}{2!\ 3!}$

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Question 195 Marks

How many different words can be formed from the letters of the word $\text{'GANESHPURI'}?$ In how many of these words:

The letter $G$ always occupies the first place?
The letters $P$ and $I$ respectively occupy first and last place?
The vowels are always together?
The vowels always occupy even places?

Answer

There are $10$ letters in the word $\text{'GANESHPURI'.}$ The total number of words formed is equal to $^{10}P_{10} = 10!$

If we fix up $G$ in the begining, then the remaining $9$ letters can be arranged in $^9P_9 = 9!$ ways
If we fix up $P$ in the begining and $I$ at the end, begining $8$ letters can be arranged in $^8P_8 = 8!.$
There are $4$ vowels and $6$ consonants in the word $\text{'GANESHPURI'.}$

Considening $4$ vowels as one letter,
We have $7$ letters which can be arranged in $^7P_7 = 7!$ ways.
$\text{A, E, U, I}$ can be put together in $4!$ ways.
Hence, required number of words $= 7! \times 4!.$

We have to arrange $10$ letters in a row such that vowels occupy even places. There are $5$ even places $(2, 4, 6, 8, 10).\ 4$ vowels can be arranged in these $5$ even places in $^5p_4$ ways.

Remaining $5$ odd places $(1, 3 , 5, 7, 9)$ are to be occupied by the $6$ consonants.
This can be done in $^6C_5$ ways.
Hence, the total number of words in which vowels occupy even places $=\ ^5\text{P}_4 \times \ ^6\text{P}_5$
$=\frac{5!}{(5-4)!}\times \frac{6!}{(6-1)!}$
$=5!\times 6!$

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Question 205 Marks

The letters of the word 'SURITI' are written in all possible orders and these words are written out as in a dictionary. Find the rank of the word 'SURITI'.

Answer

In a dictionary the words at each stage are arranged in alphabetical order.
Starting with letter I, and arranging the other 5 letters, we obtain 5!= 5 × 4 × 3 × 2 × 1 = 120. Then starting with R, and arranging the other five letters I, I, S, T, U in different ways, we obtain $=\frac{5!}{2!}=\frac{120}{2}=60.$
Number of words beginning with 5 is $\frac{5!}{2!}$, but one of these words is the word SURITI itself. So, we first find the number of words beginning with SI, SR, ST, SUI and SURI.
Number of words starting with SI $=4!=24$
Number of words starting with SR $=\frac{4!}{2!}=12$
Number of words starting with ST $=\frac{4!}{2!}=12$
Number of words starting with SUI $=3!=6$
Now, the words beginning with 'SUR' must follow.
There are $\frac{3!}{2!}=3$ words beginning with SUR one of these words is the word SURITI.
The first word beginning which SUR is the word SURIIT and the next word is SURITI.
$\therefore$ Rank of SURITI
= 120 + 60 + 24 + 2 × 12 + 6 + 2
= 180 + 56
= 236

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Question 215 Marks

Find the number of numbers, greater than a million, that can be formed with the digits 2, 3, 0, 3, 4, 2, 3.

Answer

Total num er of digits = 7 Since, 0 cannot be first digit of the 7 digit numbers. $\therefore$ Number of 6 - digitNumbers $=\frac{6!}{2!\ 3!}$ $\begin{bmatrix}\because 2\text{ comes} \ \text{2 times and 3 comes 3 times}\end{bmatrix}\\$
$=\frac{6\times5\times4\times3!}{2\times3!}$ $=6\times5\times2$ $=60.$ Now, number of 7-digit numbers $=\frac{7!}{2!\ 3!} = \frac{7\times6\times5\times4\times3!}{2\times3!}$ $=7\times6\times5\times2$ $=420$Hence, total number of numbers which is greater then 1 million= 420 - 60
= 360.

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Question 225 Marks

Find the total number of arrangements of the letters in the expression $a^3 b^2 c^4 $ when written at full length.

Answer

There are $3a's, 2b's$ and $4c's.$
So, the number of arrangements
$=\frac{9!}{4!\ 3!\ 2!}$
$=\frac{9\times8\times7\times6\times5\times4!}{4!\times3\times2\times2}$
$= 9\times4\times7\times5$
$=1260.$

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Question 235 Marks

How many words can be formed with the letters of the word 'UNIVERSITY', the vowels remaining together?

Answer

In the word 'UNIVERSITY' there are 10 letters of which 2 are I's. There are 4 vowels in the given word of which 2 are I's.
These vowels can be put together in $\frac{4!}{2!}$ ways.
Considering these 4 vowels as one letter there are 7 letters which can be arranged in 7! ways. Hence, by fundamental principle of multiplication, the required number of arrangements is
$=\frac{4!}{2!}\times 7!$
$=\frac{4\times3\times2!}{2!}\times7\times6\times5\times4\times3\times2$
$= 4\times3\times7\times6\times5\times4\times3\times2$\
$=60480$

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Question 245 Marks

If $^\text{n+5}\text{P}_\text{n+1}=-\frac{11(\text{n-1})}{2}\ ^\text{n+3}\text{P}_\text{n}$ find n.

Answer

We have,
$^\text{n+5}\text{P}_\text{n+1}=-\frac{11(\text{n-1})}{2}\ ^\text{n+3}\text{P}_\text{n}$
$\Rightarrow \frac{(\text{n+5)}!}{[\text{n+5}-(\text{n+1})]!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{[\text{n}+3-\text{n}]!}$
$\Rightarrow \frac{(\text{n+5)}!}{[\text{n+5}-\text{n-1}]!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)}{3!}$
$\Rightarrow \frac{(\text{n+5)}!}{4!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{3!}$
$\Rightarrow \frac{(\text{n+5)}(\text{n+4)}(\text{n+3)}!}{4!}=\frac{11(\text{n-1})}{2}\times\frac{(\text{n}+3)!}{3!}$
$\Rightarrow \frac{(\text{n+5)}(\text{n+4)}!}{4\times3!}=\frac{11(\text{n-1})}{2\times3}$
$\Rightarrow (\text{n}+5)(\text{n}+4)=\frac{11(\text{n}-1)\times4}{2}$
$\Rightarrow (\text{n}+5)(\text{n}+4)=22(\text{n}-1)$
$\Rightarrow \text{n}^2+4\text{n}+5\text{n}+20=22\text{n}-22$
$\Rightarrow \text{n}^2+9\text{n}-22\text{n}+20+22=0$
$\Rightarrow \text{n}^2-13\text{n}+42=0$
$\Rightarrow \text{n}^2-6\text{n}-7\text{n}+42=0$
$\Rightarrow \text{n}(\text{n}-6)-7(\text{n}-6)=0$
$\Rightarrow \text{n}=6 \ \text{or},\ \text{n}=7$
Hence, $\text{n} =6\ \text{or},\ 7$

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Question 255 Marks

If P(n, 5) : P(n, 3) = 2 : 1, find n.

Answer

We have,P(n, 5) : P(n, 3) = 2 : 1
$\Rightarrow \frac{\text{p}(\text{n},5)}{\text{p}(\text{n},3)}=\frac{2}{1}$ $\Rightarrow\frac{\frac{\text{(n)!}}{(\text{(n-5)!})}}{\frac{\text{n}!}{\text{(n-3)!}}}=\frac{2}{1}$ $\Rightarrow \frac{\text{n}!\times\text{(n-3)!}}{\text{(n-5)!}\times\text{n}!}=2$ $\Rightarrow \frac{\text{(n-3)!}}{\text{(n-5)}!}=2$ $\Rightarrow \frac{(\text{n}-3)(\text{n}-4)(\text{n}-5)!}{(\text{n}-5)!}= 2$ $\Rightarrow (\text{n}-3)(\text{n}-4) =2$ $\Rightarrow \text{n}^2-4\text{n}-3\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12-2=0$ $\Rightarrow \text{n}^2+7\text{n}+10=0$ $\Rightarrow \text{n}^2-5\text{n}-2\text{n}+10=0$ $\Rightarrow \text{n}(\text{n}-5)-2(\text{n}-5)= 0$ $\Rightarrow (\text{n}-5)(\text{n}-2) =0$ $\Rightarrow \text{n}=5 \ \begin{bmatrix}\ \because\text{r}\ \geq\ 5 \\ \therefore \ \neq\ 2 \end{bmatrix}$ Hence, $\text{n}=5$

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Question 265 Marks

How many number of four digits can be formed with the digits 1, 3, 3, 0?

Answer

Total number of digits = 4
Total number of 4 digit numbers $=\frac{4!}{2!}$
But, zero cannot be first digit of the four digit numbers.
$\therefore$ Total number of 3 digit numbers $=\frac{3!}{2!}$
$\therefore$ Total number of numbers $= \frac{4!}{2!}-\frac{3!}{2!}$
$=\frac{4\times3\times2!}{2!}-\frac{3\times2!}{2!}$
$=12-3$
$=9$
Hence, total number of four digit numbers = 9.

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Question 275 Marks

How many permutations of the letters of the word 'MADHUBANI' do not begin with M but end with I?

Answer

MADHUBANI Total number of words that ends with letter $\text{I}=\frac{8!}{2!}$ $=8\times7\times6\times5\times4\times3$ $=50\times30\times12$ $=20160$If the words starts with Mand end with I, there are 7 space left for 7 letters.
Number of words that starts with M and end with $\text{I}=\frac{7!}{2!}$
$=7\times5\times4\times3$ $=42\times60$ $=2520$Number of words which do not start with M but end with I
$= 20160 - 2520$ $= 17640$ Required number of words = 17640

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Question 285 Marks

Find the number of words formed by permuting all the letters of the following words:
RUSSIA.

Answer

There are 6 letters in the word 'RUSSIA' out of which 2 are S's, and the rest are all distinct.
So, the total number of words
$=\frac{6!}{2!\ }$
$=\frac{6\times5\times4\times3\times2!}{2!}$
$=6\times5\times4\times3$
$=360$

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Question 295 Marks

How many different signals can be made from 4 red, 2 white and 3 green flags by arranging all of them vertically on a flagstaff?

Answer

Total number of red flags = 4 Total number of white flags = 2 Total number of green flags = 3 We have to arrange 9 flags, out of which 4 are of red, 2 are white and 3 are green, So, total number of signals $= \frac{ 9!}{ 4!\ 2!\ 3!}$ $=\frac{9\times8\times7\times6\times5\times4!}{4!\times2\times3\times2}=9\times4\times7\times5=1260$Hence, total number of signals = 1260.

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Question 305 Marks

If P(15, r − 1) : P(16, r − 2) = 3 : 4, find r.

Answer

We have,
P(15, r − 1) : P(16, r − 2) = 3 : 4
$\Rightarrow \frac{\text{p}(15,\text{r}-1)}{\text{p}(16,\text{r}-2)}=\frac{3}{4}$
$\Rightarrow\frac{\frac{15!}{[15-(\text{r}-1)]!}}{\frac{16!}{[16-(\text{r}-2)]!}}=\frac{3}{4}$
$\Rightarrow\frac{\frac{15!}{[16-\text{r}]!}}{\frac{16!}{[18-\text{r}]!}}=\frac{3}{4}$
$\Rightarrow\frac{15!}{(16-\text{r})!}\times\frac{(18-\text{r})!}{16!}=\frac{3}{4}$
$\Rightarrow \frac{15\times (18-\text{r})(17-\text{r})(16-\text{r})!}{(16-\text{r})!\times16\times15!}=\frac{3}{4}$
$\Rightarrow \frac{(18-\text{r})(17-\text{r})}{16}=\frac{3}{4}$
$\Rightarrow 306-18\text{r}-17\text{r}+\text{r}^2=\frac{3}{4}\times16$
$\Rightarrow \text{r}^2-35\text{r}+306=12$
$\Rightarrow \text{r}^2-35\text{r}+306-12=0$
$\Rightarrow \text{r}^2-35\text{r}+294=0$
$\Rightarrow \text{r}^2-21 \text{r}-14 \text{r}+294=0$
$\Rightarrow \text{r}( \text{r}-21)-14( \text{r}-21)=0$
$\Rightarrow ( \text{r}-21)( \text{r}-14)=0$
$\Rightarrow \text{r}-14=0$
$\Rightarrow \text{r}=14$
Hence, $ \text{r}=14$

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Question 315 Marks

Find the number of words formed by permuting all the letters of the following words:
INTERMEDIATE.

Answer

There are 12 letters in the word 'INTERMEDIATE' out of which 2 are I'S, 2 are T'S, 3 are E'S and the rest are all distinct.
so, the total number of words
$=\frac{12!}{2!\ 2!\ 3!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2!}{2!\ 2!\ 3!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3}{2\times2\times3 }$
$=11\times10\times9\times8\times6\times5\times4\times3$
$=19958400$

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Question 325 Marks

How many three letter words can be made using the letters of the word 'ORIENTAL'?

Answer

There are 8 letters in the word 'ORIENTAL'. The total number of three letter words is the number of arrangements of 8 items, taken 3 at a time, which is equal to $=\ ^8\text{P}_3$$=\frac{8!}{(8-3)!}$
$=\frac{ 6\times 5\times4\times3\times2\times1}{5!}$
$=336$
Hence, the total number three letter words are 336.

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Question 335 Marks

Find the number of words formed by permuting all the letters of the following words:
PAKISTAN.

Answer

There are 8 letters in the word 'PAKISTAN' out of which 2 are A'S, and the rest are all distinct.
So, the total number of words
$=\frac{8!}{2!\ }$
$=\frac{8\times7\times6\times5\times4\times3\times2!}{2!}$
$=8\times7\times6\times5\times4\times3$
$=20160$

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Question 345 Marks

If the letters of the word 'MOTHER' are written in all possible orders and these words are written out as in a dictionary, find the rank of the word 'MOTHER'.

Answer

In the dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with E, H, M, 0, R, T in order. E will occur in the first place as often as there are ways of arranging the remaining 5 letters,
$\therefore$ Number of words starting with E = 5! = 5 × 4 × 3 × 2 × 1 = 120
Number of words starting with H = 5! = 120.
Number of words beginning with M is 5!, but one of these words is the word MOTHER.
So, we first find the number of words beginning with ME and MH.
Number of words starting with ME = 4! = 4 × 3 × 2 × 1 = 24.
Now, the words beginning with 'MO' must follow.
There are 4! words beginning with MO, one of these words is the word MOTHER itself.
So, we first find the number of words beginning with MOE, MOH and MOR.
Number of words starting with MOE = 3! = 6
Number of words starting with MOH = 3! = 6
Number of words starting with MOR = 3! = 6
Number of words beginning with MOT is 3! but one of these words is the word MOTHER itself
So, we first find the number of words beginning with MOTE.
Number of words starting with MOTE = 2! = 2
Now, the words beginning with MOTH must follow.
There are 2! words beginning with MOTH, one of these words is word MOTHER itself.
The first word beginning with MOTH is the word MOTHER.
$\therefore$ Rank of MOTHER = 2 × 120 + 2 × 24 + 3 × 6 + 2 + 1
= 240 + 48 + 18 + 3
= 309

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Question 355 Marks

The letters of the word 'ZENITH' are written in all possible orders. How many words are possible if all these words are written out as in a dictionary? What is the rank of the word 'ZENITH'?

Answer

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with E, H, I, N, T, Z in order. 'E' will occur in the first place as often as there are ways of arranging the remaining 5 letters all at a time i.e. E will occur S! times. Similarly H will occur in the first place the same number of times. $\therefore$ Number of words starting with E = 5! = 5 × 4 × 3 × 2 × 1 = 120Number of words starting with H = 5! = 120
Number of words starting with I = 5! = 120
Number of words starting with N = 5! = 120
Number of words starting with T = 5! = 120
Number of words beginning with Z is S!, but one of these words is the word ZEN ITH i tselr. So, we first find the number of words beginning with ZEH, ZEI and ZENH,
Number of words starting with ZEH = 3! = 6
Number of words starting with ZEI = 3! = 6
Number of words starting with ZENH = 2! = 2.
Now, the words beginning with ZENI must follow.
There are 21 words beginning with ZENI one of these words is the word ZENITH itself. The first word beginning with ZENI is the word ZENI HT and the next word is ZENITH.
$\therefore$ Rank of ZENITH = 5 × 120 + 2 × 6 + 2 + 2
= 600 + 12 + 4
= 600 + 16
= 616

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Question 365 Marks

In how many ways can the letters of the word $\text{"INTERMEDIATE"}$ be arranged so that:

The vowels always occupy even places?
The relative order of vowels and consonants do not alter?

Answer

$\ce{INTERMEDIATEI = 2 times, T = 2 times, E = 3 times, N, R, M, D, A}$
Number of letters $= 12$

There are $6$ vowels. They ocuepy even places $2^{nd}, 4^{th}, 6^{th}, 8^{th}, 10^{th}, 12^{th}.$ After there six there are six places and $5$ letters, $T$ is $2$ times. So, number of ways for consonants $=\frac{6!}{2!}$

The total number of ways when vowels ocuepy even places
$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=\frac{6\times5\times4\times3\times2\times6\times5\times4\times3\times2}{2\times2\times3\times2}$
$=21600$
Required number of ways$= 21600$

Number of ways such that relative order of vowels and consonants do not alter

$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=21600$
Required number of ways$= 21600$

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Question 375 Marks

If the permutations of a, b, c, d, e taken all together be written down in alphabetical order as in dictionary and numbered, find the rank of the permutation debac.

Answer

In a dictionary the words at each stage are arranged in alphabetical order. In the given problem we must therefore consider the words beginning with a, b, c, d, e in order. 'a' will occur in the first place as often as there are ways of arranging the remaining 4 letters all at a time i.e 'a' will occur 4! times. similarly b and c will occur in the first place the same number of times,
$\therefore$ Number of words starting with'a' = 4! = 4 × 3 × 2 × 1 = 24
Number of words starting with 'b' = 4! = 4 × 3 × 2 × 1 = 24
Number of words starting with 'c' = 4! = 4 × 3 × 2 × 1 = 24
Number of words beginning with 'd' is 4!, but one of these words is the word debac. So, we first find the number of words beginning with da, db, de, and dea
Number of words starting with da = 3! = 6
Number of words starting with db = 3! = 6
Number of words starting with de = 3! = 6
Number of words starting with dea = 2! = 2
There are 2! words beginning with deb one of these words is the word debac itself .
The first word beginning with deb is the word debac.
$\therefore$ Rank of debac = 3 × 24 + 3 × 6 + 2 + 1
= 72 + 18 + 3
= 90 + 3
= 93

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