MCQ 11 Mark
The number of ways in which the letters of the word $\text{'CONSTANT'}$ can be arranged without changing the relative positions of the vowels and consonants is.
AnswerThe word $\text{CONSTANT}$ consists of two vowels that are placed at the $2^{nd}$ and $6^{th}$ position, and six consonants.
The two vowels can be arranged at their respective places,
i.e. $2^{nd}$ and $6^{th}$ place, in $2!$ ways.
The remaining $6$ consonants can be arranged at their respective places in $\frac{6!}{2!\ 2!}$ ways.
$\therefore$ Total number of arrangements $=2!\times\frac{6!}{2!\ 2!}=360$
View full question & answer→MCQ 21 Mark
The number of non$-$negative integral solutions of $\text{x} + \text{y}+\text{z}\leq\text{n},$ where $\text{n}\in\text{N}$ is:
- ✓
$\text{n}+\ ^3\text{C}_3$
- B
$\text{n}+\ ^4\text{C}_4$
- C
$\text{n}+\ ^5\text{C}_5$
- D
$\text{n}+\ ^2\text{C}_2$
AnswerCorrect option: A. $\text{n}+\ ^3\text{C}_3$
View full question & answer→MCQ 31 Mark
How many $3 -$ letter words with or without meaning, can be formed out of the letters of the word, $\text{LOGARITHMS,}$ if repetition of letters is not allowed:
AnswerThe word $\text{LOGARITHMS}$ has $10$ different letters.
Hence, the number of $3-$letter words $($with or without meaning$)$ formed by using these letters.
$=\ ^{10}\text{P}_3$
$=10\times9\times8$
$=720$
View full question & answer→MCQ 41 Mark
In a chess tournament each of six players will play every other player exactly once. How many matches will be played during the tournament?
AnswerFirst player can play $5$ matches with other five players.
Second player can play $4$ matches with other four players and proceeding this way, the fifth player will playonly one match with sixth player.
$\therefore$ Total number of matches played $= 5 +4 + 3 + 2 + 1 = 15.$
Short cut : Number of matches played by $4$ players $=\frac{\text{n}(\text{n+1)}}{2}$
View full question & answer→MCQ 51 Mark
In a crossword puzzle, $20$ words are to be guessed of which $88$ words have each an alternative solution also.The number of possible solutions will be:
- A
$\ ^{20}\text{P}_8$
- B
$\ ^{20}\text{C}_8$
- C
$515$
- ✓
$256$
Answer$8$ out of the $20$ words have an alternative solution.
So $12$ words are fixed,
so they wont affect the number of ways.
So now we find the number of ways in which the remaining $8$ words can be guessed.
Now every word have $2$ solutions,
so there are $2$ ways to fill each of these words, and for $8$ such words, it can be done in
$2\times2\times2\times2\times2\times2\times2\times2$
$=2^8$
$=256$
View full question & answer→MCQ 61 Mark
How many factors are $2^5\times3^6\times5^2$ are perfect squares:
AnswerAny factors of $ 2^5\times3^6\times5^2 $ which is a perfect square will be of the form $ 2^\text{a}\times3^\text{b}\times5^\text{c} $
where a can be $0$ or $2$ or $4,$
So there are $3$ ways.
$b$ can be $0$ or $2$ or $4$ or $6$,
So there are $4$ ways.
$a$ can be $0$ or $2,$
So there are $2$ ways.
So, the required number of factors $=3\times4\times2=24$
View full question & answer→MCQ 71 Mark
There are $44$ candidates for a Natural science scholarship, $22$ for a Classical and $66$ for a Mathematical scholarship, then find the no.of ways one of these scholarship can be awarded is:
AnswerNatural science scholarship can be awarded to anyone of the four candidates.
So, there are $44$ ways of awarding natural science scholarships.
Similarly, mathematical and classical scholarships can be awarded in $22$ and $66$ ways respectively.
By the fundamental principle of addition, number of ways of awarding one of the three scholarships is.
$= 4 + 2 + 6$
$= 12$
View full question & answer→MCQ 81 Mark
If $^\text{k}+5\text{P}_\text{k+1}=\frac{11(\text{k}-3)}{2}.\ ^\text{k+3}\text{P}_\text{k}$, then the values of $k$ are:
- A
$7$ and $11$
- ✓
$6$ and $7$
- C
$2$ and $11$
- D
$2$ and $6$
AnswerCorrect option: B. $6$ and $7$
$^\text{k}+5\text{P}_\text{k+1}=\frac{11(\text{k}-3)}{2}.\ ^\text{k+3}\text{P}_\text{k}$
$\Rightarrow \frac{(\text{k}+5)!}{(\text{k}+5-\text{k}-1)!}=\frac{11(\text{k}-1)}{2}\times \frac{(\text{k}+3)!}{(\text{k}+3-\text{k})!}$
$\Rightarrow \frac{(\text{k}+5)!}{(\text{k}+3)!}=\frac{11(\text{k}-1)}{2}\times\frac{4!}{3!}$
$\Rightarrow (\text{k}+5)(\text{k}+4)=22(\text{k}-1)$
$\Rightarrow \text{k}^2+9\text{k}+20=22\text{k}-22$
$\Rightarrow \text{k}^2-13\text{k}+42=0$
$\Rightarrow \text{k}=6,7$
View full question & answer→MCQ 91 Mark
The number of five$-$digit telephone numbers having at least one of their digits repeated is:
- A
$90000.$
- B
$100000.$
- C
$30240.$
- ✓
$69760$
AnswerCorrect option: D. $69760$
Total number of five digit numbers $($since there is no restriction of the number $0XXXX)$
$= 10 \times 10 \times 10 \times 10 \times 10$
$= 100000.$
These numbers also include the numbers where the digits are not being repeated.
So, we need to subtract all such numbers.
Number of $5$ digit numbers that can be formed without any repetition of digits
$= 10 \times 9 \times 8 \times 7 \times 6$
$= 30240$
$\therefore$ Number of five$-$digit telephone numbers having at least one of their digits repeated
$= \{$Total number of $5$ digit numbers$\} - \{$Number of numbers that do not have any digit repeated$\}$ $= 100000 - 30240$
$= 69760$
View full question & answer→MCQ 101 Mark
The number of rectangles that you can find on a chess board is:
AnswerCorrect option: B. $1296$
View full question & answer→MCQ 111 Mark
The number of ways in which the letters of the word $\text{ARTICLE}$ can be arranged so that even places are always occupied by consonants is:
- ✓
$576$
- B
$^4C_3 \times 4!$
- C
$2 \times 4!$
- D
AnswerThere are $3$ even places in the $7$ letter word $\text{ARTICLE.}$
So, we have to arrange $4$ consonants in these $3$ places in $\ ^4P_3$ ways.
And the remaining $4$ letters can be arranged among themselves in $4!$ ways.
$\therefore$ Total number of ways of arrangement $=\ ^4P_3\times 4! = 4! \times 4! = 576$
View full question & answer→MCQ 121 Mark
It is required to seat $5$ men and $4$ women in a row so that the women occupy the even places.The number of ways such arrangements are possible are
- A
$8820$
- ✓
$2880$
- C
$2088$
- D
$2808$
AnswerCorrect option: B. $2880$
Total number of persons are $9$ in which there are $5$ men and $4$ women.
So total number of place $= 9$
Now women seat in even place.
So total number of arrangement $\text{= 4! (_W_W_W_W_) (W-Woman)}$
Men sit in odd place.
So total number of arrangement $= 5! (\text{MWMWMWMWM) (M-Man)}$
Now Total number of arrangement $= 5! \times 4! = 120 \times 24 = 2880$
View full question & answer→MCQ 131 Mark
Total number of four digit odd numbers that can be formed using $0, 1, 2, 3, 5, 7 ($using repetition allowed$)$ are:
View full question & answer→MCQ 141 Mark
In how many ways a committee consisting of $5$ men and $3$ women, can be chosen from $9$ men and $12$ women:
- A
$10258$
- B
$16870$
- ✓
$27720$
- D
$38982$
AnswerCorrect option: C. $27720$
View full question & answer→MCQ 151 Mark
Choose the correct option for the following. $n! = n(n − 1)(n − 2).....3.2.1:$
AnswerFactorial: the product of an integer and all integer less than that
$\therefore n! = n \times (n − 1) \times (n − 2).....3 \times 2 \times 1$
$\therefore$ The given statement
$n! = n \times (n − 1) \times (n − 2)....3 \times 2 \times 1$ is True.
View full question & answer→MCQ 161 Mark
If $ (1 + \text{x})^\text{ⁿ} = \text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}ⁿ,$ then the value of$\ ^\text{C}0^² + \ ^\text{C}1^² + \ ^\text{C}2^² + .....+ ^\text{C}\text{n}^\text{ⁿ} =\ ^{2\text{n}}\text{C}_\text{n}$ is:
- A
$\frac{(2\text{n})!}{(\text{n}!)}$
- ✓
$\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)}$
- C
$\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)2}$
- D
$\text{None of these}$
AnswerCorrect option: B. $\frac{(2\text{n})!}{(\text{n}!\times\text{n}!)}$
Given,$ (1 + \text{x})^\text{ⁿ} = \text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}ⁿ .....1$
and $(1 + \text{x})^\text{n} = \ ^\text{C}0 \text{ x}^\text{n} + \ ^\text{C}1\text{x}^\text{n-1}+ \ ^\text{C}2 \text{ x}^\text{n-2} + .... \ ^\text{C}\text{r}\text{x }^\text{n-r} + … + \ ^\text{C}\text{n-1 } \text{x} + \ ^\text{C}\text{n } ... 2$
Multiply $1$ and $2,$ we get
$ (1 + \text{x})^\text{2ⁿ} = (\text{C}_{0} +\text{C}_1\text{x} +\text{C}_2\text{ x}^² + …+ \ ^\text{C}\text{n} \text{ x}^\text{n}\times)(\ ^\text{C}0\text{ x}^\text{n}+\text{C}_1\text{ x}^\text{n-1}+\text{C}_2\text{ x}^\text{n-2}+...+\text{C}_\text{r} \text{x}^\text{n-r}+...+\text{C}_\text{n-1}\text{ x}+\text{C}_\text{n})$
Now, equating the coefficient of $xn$ on both side, we get
$\ ^\text{C}0^2 + \ ^\text{C}1^2 + \ ^\text{C}2^² + ….+ \ ^\text{C}\text{n}^\text{n} = \ ^\text{2n}\text{C}_\text{n} = \frac{(2\text{n})!}{(\text{n}! × \text{n}!)}$
View full question & answer→MCQ 171 Mark
There are mn letters and $n$ post boxes.The number of ways in which these letters can be posted is:
- A
$(\text{mn})^\text{n}$
- B
$(\text{mn})^\text{m}$
- C
$\text{m}^\text{mn}$
- ✓
$\text{n}^\text{mn}$
AnswerCorrect option: D. $\text{n}^\text{mn}$
Every letter can be posted in any of the n post boxes.
$\therefore$ Total number of ways $=\text{n}\times\text{n}\times\text{n}\times.....(\text{m}\times\text{n})=\text{n}^\text{mn}$
View full question & answer→MCQ 181 Mark
In how many ways in which $8$ students can be sated in a circle is:
- A
$40302$
- B
$40320$
- ✓
$5040$
- D
$50040$
AnswerCorrect option: C. $5040$
The number of ways in which $8$ students can be sated in a circle $= ( 8 – 1)!$
$= 7!$
$= 5040$
View full question & answer→MCQ 191 Mark
A car driver knows four different routes from Delhi to Amritsar. From Amritsar to Pathankot, he knows three different routes and from Pathankot to Jammu he knows two different routes. How many routes does he know from Delhi to Jammu?
AnswerThe car driver can reach Amritsar in $4$ ways.
From each of these four ways, he can reach Pathankot in $3$ different ways and so he can reach from Delhi to Pathankot in $(4 \times 3)$
i.e $12$ ways.
Again from Pathankot to Jammu, there are $2$ ways.
Hence he can reach Jammu from Delhi in $(12 \times 2)$ i.e. in $24$ ways.
View full question & answer→MCQ 201 Mark
A $5-$digit number divisible by $3$ is to be formed using the digits $0, 1, 2, 3, 4$ and $5$ without repetition. The total number of ways in which this can be done is:
- ✓
$216$
- B
$600$
- C
$240$
- D
$3125$
AnswerA number is divisible by $3$ when the sum of the digits of the number is divisible by $3.$
Out of the given $6$ digits,
there are only two groups consisting of $5$ digits whose sum is divisible by $3.$
$= 1 + 2 + 3 + 4 + 5 = 15$
$= 0 + 1 + 2 + 4 + 5 = 12$
Using the digits $1, 2, 3, 4$ and $5,$ the $5$ digit numbers that can be formed $= 5!$
Similarly, using the digits $0, 1, 2, 4$ and $5,$
the number that can be formed $= 5! - 4! \{$since the first digit cannot be $0\}$
$\therefore$ Total numbers that are possible $= 5! + 5! - 4! = 240 - 24 = 216$
View full question & answer→MCQ 211 Mark
Six identical coins are arranged in a row.The number of ways in which the number of tails is equal to the number of heads is:
MCQ 221 Mark
Choose the correct answer. Total number of words formed by $2$ vowels and $3$ consonants taken from $4$ vowels and $5$ consonants is equal to.
AnswerCorrect option: C. $7200$
Given that total number of vowels $= 4$
and number of consonants $= 5$
The total of words formed by $2$ vowels and $3$ consonents
$=\ ^4\text{C}_2\times\ ^5\text{C}_3$
$=\frac{4!}{2!\ 2!}\times\frac{5!}{3!\ 2!}$
$=\frac{4\times3\times2!}{2\times1\times2!}\times\frac{5\times4\times3!}{3!\times2}$
Now permutation of $2$ vowels and $3$ consonants $= 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$
So, the total number of words $= 60 \times 120 = 7200.$
View full question & answer→MCQ 231 Mark
A group of $1200$ persons consisting of captains and soldiers is travelling in a train. For every $15$ soldiers there is one captain. The number of captains in the group is:
AnswerOut of $16$ men, there is a captain.
Number of captains in $1200$ men $= 1200 \div 16 = 75.$
View full question & answer→MCQ 241 Mark
If a secretary and a joint secretary are to be selected from acommittee of $11$ members, then in how many ways can they be selected:
View full question & answer→MCQ 251 Mark
An automobile dealer provides motorcycles and scooters in three body patterns and $4$ different colors each. The number of choices open to a customer is:
- A
$5C_3$
- B
$4C_3$
- C
$4 \times 3$
- ✓
$4 \times 3 \times 2$
AnswerCorrect option: D. $4 \times 3 \times 2$
View full question & answer→MCQ 261 Mark
There are $15$ points in a plane, no two of which are in a straight line except $4,$ all of which are in a straight line.The number of triangle that can be formed by using these $15$ points is:
- A
$\ ^{15}\text{C}_3$
- B
$490$
- ✓
$451$
- D
$415$
AnswerThe required number of triangle $ =\ ^{15}\text{C}_3-\ ^4\text{C}_3 = 455-4 = 451$
View full question & answer→MCQ 271 Mark
Find the number of rectangles and squares in an $8$ by $8$ chess board respectively.
- A
$296, 204$
- ✓
$1092, 204$
- C
$204, 1092$
- D
$204, 1296$
AnswerCorrect option: B. $1092, 204$
Chess board consists of $9$ horizontal $9$ vertical lines.
A rectangle can be formed by any two horizontal and two vertical lines.
Number of rectangles $= \ ^9\text{C}_2\times\ ^9\text{C}_2 = 1296.$
For squares there is one $8$ by $8$ square four $7$ by $7$ squares, nine $6$ by $6$ squares and like this
Number of squares on chess board $= 1^2+2^2…..8^2 = 204$
Only rectangles $ = 1296-204 = 1092$
View full question & answer→MCQ 281 Mark
Six boys and six girls sit along a line alternately in $x$ ways and along a circle $($again alternatively in $y$ ways$)$, then:
- A
$x = y$
- B
$y = 12x$
- C
$x = 10y$
- ✓
$x = 12y$
AnswerCorrect option: D. $x = 12y$
Given, six boys and six girls sit along a line alternately in $x$ ways and along a circle
$($again alternatively in $y$ ways$).$
Now, $\text{x} = 6!\times6!+6!\times6!$
$\Rightarrow\text{x}=2\times(6!)^2$
and $\text{y}=5!\times6!$
Now,$\frac{\text{x}}{\text{y}} =\frac{{2\times(6!)^2}}{(5!\times6!)}$
$\Rightarrow\frac{\text{x}}{\text{y}}=\frac{(2\times6!\times6!)}{5!\times6\!}$
$\Rightarrow\frac{\text{x}}{\text{y}} =\frac{({2\times6!})}{5!}$
$\Rightarrow\frac{\text{x}}{\text{y}} =\frac{({2\times6\times5!})}{5!}$
$\frac{\text{x}}{\text{y}} = 12$
$\Rightarrow\text{x} = 12\text{y}$
View full question & answer→MCQ 291 Mark
Out of $100$ students $50$ fail in English and $30$ in Maths. If $12$ students fail in both English and Maths, then the number of students passing both the subjects is:
AnswerTotal number of students $= 100$
Number of students fail in English $= (50 - 12) = 38$
Number of students fail in Maths $= (30 - 12) = 18$
Number of students fail in both $= 12$
Therefore total failing students $= (38 + 18 + 12) = 68$
Pass in both the subjects $= (100 - 68) = 32$ students
$32$ students have passed in both subjects.
View full question & answer→MCQ 301 Mark
$^xC_7 −^xC_5 = 0$, then $x =:$
View full question & answer→MCQ 311 Mark
If $\ ^\text{n}\text{P}_5 = 60\ ^\text{n-1}\text{P}_3,$ the value of $n$ is:
AnswerGiven that $\ ^\text{n}\text{P}_5 = 60\ ^\text{n-1}\text{P}_3,$
We know that $\text{P}(\text{n, r}) = \ ^\text{n}\text{P}_r = \frac{\text{n!}}{(\text{n-r})!}$
Now, apply the formula on both sides to get the value of $n.$
$\frac{\text{n!}}{(\text{n}-5)!} = 60 \bigg[\frac{(\text{n}-1)!}{(\text{n}-1)-3!}\bigg]$
On solving the above equation,
we get $n= -6$ and $n=10.$
Since the value of $n$ cannot be negative, the value of $n$ is $10.$
View full question & answer→MCQ 321 Mark
Arranging people, digits, numbers, alphabets, letters, and colours are example of:
AnswerPermutation : Arranging people, digits, numbers, alphabets, letters, and colours:
View full question & answer→MCQ 331 Mark
There are $4$ parcels and $5$ post offices.In how many ways can $4$ parcels be got registered:
- A
$20$
- B
$4^5$
- ✓
$5^4$
- D
$5^4- 4^5$
View full question & answer→MCQ 341 Mark
The number of $6 -$ digit numbers can be formed from the digits $0, 1, 3, 5, 7$ and $9$ which are divisible by $10$ and no digit is repeated are:
AnswerA number is divisible by $10$ if the unit digit of the number is $0.$
Given digits are $0, 1, 3, 5, 7, 9$
Now we fix digit $0$ at unit place of the number.
Remaining $5$ digits can be arranged in $5!$ ways
So, total $6 -$ digit numbers which are divisible by $10 = 5! = 120$
View full question & answer→MCQ 351 Mark
The number of different signals which can be given from $6$ flags of different colours taking one or more at a time, is:
AnswerCorrect option: B. $1956$
Number of permutations of six signals taking $1$ at a time $=\ ^6P_1$
Number of permutations of six signals taking $2$ at a time $=\ ^6P_2$
Number of permutations of six signals taking $3$ at a time $=\ ^6P_3$
Number of permutations of six signals taking $4$ at a time $=\ ^6P_4$
Number of permutations of six signals taking $5$ at a time $=\ ^6P_5$
Number of permutations of six signals taking all at a time $=\ ^6P_6$
$\therefore$ Total number of signals
$=\frac{6!}{5!}+\frac{6!}{4!}+\frac{6!}{3!}+\frac{6!}{2!}+\frac{6!}{1!}+6!$
$=6+30+120+360+720+720$
$=1956$
View full question & answer→MCQ 361 Mark
Choose the correct answer. The number of $5-$digit telephone numbers having atleast one of their digits repeated is.
- A
$90,000$
- B
$10,000$
- C
$30,240$
- ✓
$69,760$
AnswerCorrect option: D. $69,760$
Total number of telephone numbers when there is no restriction $=10^5$
Also number of telephone numbers having all digits different $={ }^{10}{P}_5$
Required number of ways
$=10^5-{ }^{10} {P}_5$
$=1000000-10 \times 9 \times 8 \times 7 \times 6$
$=1000000-30240$
$= 69760$
View full question & answer→MCQ 371 Mark
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r+4}}$ is then ${^\text{r}}\text{C}_{\text{3}}$ equal to:
Answer$\text{r}+\text{r}+4=20$
$\Rightarrow 2\text{r}+4=20$
$\Rightarrow 2\text{r}=16$
$\Rightarrow \text{r}=8$
Now,
${^\text{r}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}={^\text{8}}\text{C}_{\text{3}}$
$\therefore\ {^\text{8}}\text{C}_{\text{3}}=\frac{8!}{3!5!}$
$=\frac{8\times7\times6}{3\times2\times1}$
$=56$
View full question & answer→MCQ 381 Mark
If$\ ^\text{n+1}\text{C}_3 = 2\ ^\text{ⁿ}\text{C}_2,$ then the value of $n$ is:
AnswerGiven,$\ ^\text{n+1}\text{C}_3 = 2\ ^\text{ⁿ}\text{C}_2,$
$\Rightarrow\bigg[\frac{(\text{n + 1})!}{(\text{n} + 1 – 3)}\times3!\bigg] = \frac{2\text{n}!}{(\text{n} – 2)}\times2!$
$\Rightarrow\bigg[\frac{(\text{n}\times1!)}{(\text{n}-2)}\times3!\bigg] = \frac{2\text{n}!}{(\text{n} – 2)}\times2$
$\Rightarrow\frac{\text{n}}{3!} = 1$
$\Rightarrow\frac{\text{n}}{6} = 1$
$\Rightarrow\text{n} = 6$
View full question & answer→MCQ 391 Mark
Factorial of negative numbers is always greater than $1:$
AnswerFactorial: product of an integer with integer less than it.
Factorial can be interpolated using gamma function and gamma function and gamma function is not defined for negative integer.
Factorial is not defined for negative integer.
View full question & answer→MCQ 401 Mark
From a committee of $8$ persons, in how many ways can we choose a chairman and a vice $-$ chairman assuming one person cannot hold more than one position:
View full question & answer→MCQ 411 Mark
The number of ways in which $6$ men can be arranged in a row so that three particular men are consecutive, is:
- ✓
$4! \times 3!$
- B
$4!$
- C
$3! \times 3!$
- D
AnswerCorrect option: A. $4! \times 3!$
According to the question, $3$ men have to be 'consecutive' means that they have to be considered as a single man. But, these $3$ men can be arranged among themselves in $3!$ ways.
And, the remaining $3$ men, along with this group, can be arranged among themselves in $4!$ ways.
$\therefore$ Total number of arrangements $= 4! \times 3!$
View full question & answer→MCQ 421 Mark
The value of $({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+.....+({^\text{7}}\text{C}_{\text{6}}+{^\text{7}}\text{C}_{\text{7}})$ is:
- A
$2^{7}-1$
- ✓
$2^{8}-2$
- C
$2^{8}-1$
- D
$2^{8}$
AnswerCorrect option: B. $2^{8}-2$
$({^\text{7}}\text{C}_{\text{0}}+{^\text{7}}\text{C}_{\text{1}})+({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})+({^\text{7}}\text{C}_{\text{3}}+{^\text{7}}\text{C}_{\text{4}})+......$
$=1+2\times{^\text{7}}\text{C}_{\text{1}}+2\times{^\text{7}}\text{C}_{\text{2}}+2\times{^\text{7}}\text{C}_{\text{3}}+2\times{^\text{7}}\text{C}_{\text{4}}+2\times{^\text{7}}\text{C}_{\text{5}}..$
$=2+2^{2}({^\text{7}}\text{C}_{\text{1}}+{^\text{7}}\text{C}_{\text{2}}+{^\text{7}}\text{C}_{\text{3}})$
$=2+2^{2}(7+\frac{7}{2}\times6+\frac{7}{3}\times\frac{6}{2}\times5)$
$=2+252$
$=254$
$=2^{8}-2$
View full question & answer→MCQ 431 Mark
How many ways can $6$ coins be chosen from $20,$ one rupees coins, $10$ fifty paise coins, $7$ twenty paise coins:
- ✓
$28$
- B
$56$
- C
$\ ^{37}\text{C}_6$
- D
$38 $
View full question & answer→MCQ 441 Mark
How many $5 -$ digit telephone numbers can be constructed using the digits $0$ to $9$, if each number starts with $67$ and no digit appears more than once:
View full question & answer→MCQ 451 Mark
Three persons enter a railway compartment. If there are $5$ seats vacant, in how many ways can they take these seats?
AnswerThree persons can take $5$ seats in $^5C_3$ ways.
Moreover $3$ persons can sit in $3!$ ways.
Required number of ways ${^\text{5}}\text{C}_{\text{3}}\times3!$
$=10\times6$
$=60$
View full question & answer→MCQ 461 Mark
If ${^\text{15}}\text{C}_{3\text{r}}={^\text{15}}\text{C}_{\text{r+3}},$ is then equal to:
Answer$3\text{r}+\text{r}+3=15$
$\Rightarrow 4\text{r}+3=15$
$\Rightarrow 4\text{r}=12$
$\Rightarrow \text{r}=3$
View full question & answer→MCQ 471 Mark
Let $Tn$ denote the number of triangles which can be formed using the vertices of a regular polygon of $n$ sides.If $\text{T}_\text{n+1} -\text{T}_\text{n} = 21,$ then $n$ equals:
View full question & answer→MCQ 481 Mark
Choose the correct answer. The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is.
AnswerNumber of men $= 4$
Number of women $= 6$
We are given that the committee includes $2$ men and exactly twice as many women as men.
Thus, the possible selection can be $2$ men and $4$ women and $3$ men and $6$ women
So, the number of committee = ${ }^4 C_2 \times{ }^6 C_4+{ }^4 C_3 \times{ }^6 C_6$
$=6 \times 15+41$
$=90+4$
$=94$
View full question & answer→MCQ 491 Mark
Given $5$ flags of different colours, how many different signals can be generated, if each signal requires the use of $2$ flags, one below the other:
View full question & answer→MCQ 501 Mark
The number of numbers of $9$ different non$-$zero digits such that all the digits in the first four places are less than the digit in the middle and all the digits in the last four places are greater than the digit in the middle is:
AnswerCorrect option: B. $(4!)^2$
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