Solve the Following Question.(4 Marks)

Question 14 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
$=\frac{5!}{2!\ }$
$=\frac{5\times4\times3\times2!}{2!}$
$=60$

View full question & answer→

Question 24 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, $\text{n}=6$

View full question & answer→

Question 34 Marks

How many words (with or without dictionary meaning) can be made from the letters in the word 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

M O N D A Y 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

$=2\times\ ^5\text{P}_5$

$= 2\times5\times4\times3\times2\times1$

$=240$

View full question & answer→

Question 44 Marks

If P(11, r) = P(12, r − 1) find r.

Answer

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

View full question & answer→

Question 54 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 $=\frac{14!}{2!\ 3!\ 2!}$
$=\frac{14!}{2\times 3\times2\times2}$
$=\frac{14!}{24}$

View full question & answer→

Question 64 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

View full question & answer→

Question 74 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 $

View full question & answer→

Question 84 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
$\Rightarrow \frac{\text{p}(\text{2n-1,n})}{\text{P} (\text{2n}+1,\text{n} -1)}=\frac{22}{7}$
$\Rightarrow\frac{\frac{\text{(2n-1)!}}{(\text{(2n-1-n)!})}}{\frac{\text{(2n+1)}!}{\text{[2n+1-(n-1)]!}}}=\frac{22}{7}$
$\Rightarrow \frac{\text{(2n-1) }\times\text{(n+2)!}}{\text{(n-1)!}(\text{2n+1})!}=\frac{22}{7}$
$\Rightarrow \frac{\text{(2n-1)}\times\text{(n+2)(n+2-1(n+2-2)(n+2-3)!}}{\text{(n+1)!(2n+1).2n.(2n+1)!}}=\frac{22}{7}$
$\Rightarrow \frac{\text{n(n+2)(n+1)}}{2\text {(2n+1)}}=\frac{22}{7}$
$\Rightarrow \frac{\text{n}^2+\text{n}+\text{2n}+2}{4\text{n}+2}=\frac{22}{7}$
$\Rightarrow 7(\text{n}^2+3\text{m}+2)=22\times (4\text{n}+2)$
$\Rightarrow7\text{n}^2+21\text{n}+44=88\text{n}-44$
$\Rightarrow7\text{n}^2+21\text{n}-88\text{n}+14-44=0$
$\Rightarrow 7\text{n}^2-67\text{n}-30=0$
$\Rightarrow 7\text{n}^2 -70\text{n}+3\text{n}-30=0$
$\Rightarrow 7\text{n}(\text{n}-10)+3(\text{n}-10)=0$
$\Rightarrow (\text{n}-10)(7\text{n}+3)=0$
$\Rightarrow \text{n}-10=0$
$\Rightarrow \text{n}=10\ [\because \text{7n+3}\neq 0]$

View full question & answer→

Question 94 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
$=\frac{9!}{3!\ 2!}$
$=\frac{9\times8\times7\times6\times5\times4\times3!}{3!\times2\times1}$
$=9\times8\times7\times6\times5\times2$
$=30240$

View full question & answer→

Question 104 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 $=\frac{12!}{2!\ 3!\ 4!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4!}{2!\ 3!\ 4!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5}{2\times3\times2 }$
$=11\times10\times9\times8\times7\times6\times5$
$=1663200.$

View full question & answer→

Question 114 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.

View full question & answer→

Question 124 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, $\text{x}=12$

View full question & answer→

Question 134 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.

View full question & answer→

Question 144 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.

View full question & answer→

Question 154 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
$=\frac{7!}{2!\ 2!}$
$=\frac{7\times6\times5\times4\times3\times2!}{2!\ 2!}$
$=\frac{7\times6\times5\times4\times3}{2\times1 }$
$=7\times6\times5\times4\times3$
$=1260$

View full question & answer→

Question 164 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
$=\frac{6!}{2!\ 2!}$
$=\frac{6\times5\times4\times3\times2!}{2!\ 2!}$
$=\frac{6\times5\times4\times3}{2\times1}$
$=6\times5\times2\times3$
$=180$

View full question & answer→

Question 174 Marks

How many permutations can be formed by the letters of the word, $'VOWELS',$ 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 '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 '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 'VOWELS'.

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

View full question & answer→

Question 184 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 $\frac{9!}{2!\ 3!}$

View full question & answer→

Question 194 Marks

How many different words can be formed from the letters of the word '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 'GANESHPURI'. The total number of words formed is equal to ${ }^{10} P _{10}=10$ !
i. If we fix up $G$ in the begining, then the remaining $9$ letters can be arranged in ${ }^9 P_9=9$ ! ways
ii. If we fix up P in the begining and I at the end, begining $8$ letters can be arranged in ${ }^8 P _8=8$ !.
iii. There are $4$ vowels and $6$ consonants in the word 'GANESHPURI'.
Considening $4$ vowels as one letter,
We have $7$ letters which can be arranged in ${ }^7 P _7=7!$ ways.
A, E, U, I can be put together in $4$ ! ways.
Hence, required number of words $=7!\times 4!$.
iv. 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 ${ }^5 p_4$ ways.
Remaining $5$ odd places $(1,3,5,7,9)$ are to be occupied by the $6$ consonants.
This can be done in ${ }^6 C _5$ ways.
Hence, the total number of words in which vowels occupy even places $={ }^5 P _4 \times{ }^6 P _5$
$=\frac{5!}{(5-4)!} \times \frac{6!}{(6-1)!}$
$=5!\times 6!$

View full question & answer→

Question 204 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

View full question & answer→

Question 214 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.

View full question & answer→

Question 224 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 $3 a a^{\prime}$ s, $2 b$ 's and $4 c^{\prime}$ s.
So, the number of arrangements
$=\frac{9!}{4!\ 3!\ 2!}$
$=\frac{9\times8\times7\times6\times5\times4!}{4!\times3\times2\times2}$
$= 9\times4\times7\times5$
$=1260.$

View full question & answer→

Question 234 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
$=\frac{4!}{2!}\times 7!$
$=\frac{4\times3\times2!}{2!}\times7\times6\times5\times4\times3\times2$
$= 4\times3\times7\times6\times5\times4\times3\times2$\
$=60480$

View full question & answer→

Question 244 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, $\text{n} =6\ \text{or},\ 7$

View full question & answer→

Question 254 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, $\text{n}=5$

View full question & answer→

Question 264 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.

View full question & answer→

Question 274 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

View full question & answer→

Question 284 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
$=\frac{6!}{2!\ }$
$=\frac{6\times5\times4\times3\times2!}{2!}$
$=6\times5\times4\times3$
$=360$

View full question & answer→

Question 294 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.

View full question & answer→

Question 304 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$

View full question & answer→

Question 314 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
$=\frac{12!}{2!\ 2!\ 3!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3\times2!}{2!\ 2!\ 3!}$
$=\frac{12\times11\times10\times9\times8\times7\times6\times5\times4\times3}{2\times2\times3 }$
$=11\times10\times9\times8\times6\times5\times4\times3$
$=19958400$

View full question & answer→

Question 324 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.

View full question & answer→

Question 334 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
$=\frac{8!}{2!\ }$
$=\frac{8\times7\times6\times5\times4\times3\times2!}{2!}$
$=8\times7\times6\times5\times4\times3$
$=20160$

View full question & answer→

Maths Solve the Following Question.(4 Marks)

Take a timed test