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
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 31 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 41 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
${ }^4 C_3 \times 4$ !
- 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 ${ }^4 P_3$ ways.
And the remaining $4$ letters can be arranged among themselves in $4 !$ ways.
$\therefore$ Total number of ways of arrangement $={ }^4 p_3 \times 4!=4!\times 4!=576$
View full question & answer→MCQ 51 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 61 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 $={ }^6 P_1$
Number of permutations of six signals taking $2$ at a time $={ }^6 P_2$
Number of permutations of six signals taking $3$ at a time $={ }^6 P _3$
Number of permutations of six signals taking $4$ at a time $={ }^6 P_4$
Number of permutations of six signals taking $5$ at a time $={ }^6 P _5$
Number of permutations of six signals taking all at a time $={ }^6 P _6$
$\therefore$ 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 71 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 81 Mark
If the letters of the word $\text{KRISNA}$ are arranged in all possible ways and these words are written out as in a dictionary, then the rank of the word $\text{KRISNA}$ is :
AnswerWhen arranged alphabetically, the letters of the word $\text{KRISNA}$ are $A, I, K, N, R$ and $S$.
Number of words that will be formed with $A$ as the first letter $=$ Number of arrangements of the remaining $5$ letters $= 5!$
Number of words that will be formed with $I$ as the first letter $=$ Number of arrangements of the remaining $5$ letters $= 5!$
$\therefore$ The number of words beginning with $KA =$ Number of arrangements of the remaining $4$ letters $= 4!$
The number of words beginning with $KA =$ Number of arrangements of the remaining $4$ letters $= 4!$
The number of words starting with $KN =$ Number of arrangements of the remaining $4$ letters $= 4!$
Alphabetically, the next letter will be $KR$.
Number of words starting with $KR$ followed by $A$,
i.e. $\text{KRA} =$ Number of arrangements of the remaining $3$ letters $= 3!$
Number of words starting with $\text{KRI}$ followed by $A,$
i.e. $\text{KRIA} = $ Number of arrangements of the remaining $2$ letters $= 2!$
Number of words starting with $\text{KRI}$ followed by $N,$
i.e. $\text{KRIN} =$ Number of arrangements of the remaining $2$ letters $= 2!$
The first word beginning with $\text{KRIS}$ is the word $\text{KRISAN}$ and the next word is $\text{KRISNA}$.
$\therefore$ Rank of the word $\text{KRISNA} = 5! + 5! + 4! + 4! + 4! + 3! + 2! + 2! + 2$
$= 324$
View full question & answer→MCQ 91 Mark
If in a group of $n$ distinct objects, the number of arrangements of $4$ objects is $12$ times the number of arrangements of $2$ objects, then the number of objects is :
AnswerAccording to the question:
$^\text{n}\text{P}_4=12\times\ ^\text{n}\text{P}_2$
$\Rightarrow \frac{\text{n!}}{(\text{n}-4)!}=12\times\frac{\text{n!}}{\text{(n-2)!}}$
$\Rightarrow \frac{(\text{n}-2)!}{(\text{n}-4)!}=12$
$\Rightarrow (\text{n}-2)(\text{n}-3)=4\times 3$
$\Rightarrow \text{n}-2=4$
$\Rightarrow \text{n}=6$
View full question & answer→MCQ 101 Mark
The number of permutations of $n$ different things taking $r$ at a time when $3$ particular things are to be included is :
- A
$^{\text{n}-3}\text{P}_{\text{r}-3}$
- B
$^{\text{n}-3}\text{P}_{\text{r}}$
- C
$^{\text{n}}\text{P}_{\text{r}-3}$
- D
$\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
AnswerHere, we have to permute n things of which $3$ things are to be included.
So, only the remaining $(n − 3)$ things are left for permutation, taking $(r − 3)$ things at a time.
This is because $3$ things have already been included.
But, these $r$ things can be arranged in $r!$ ways.
$\therefore$ Total number of permutations $=\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
View full question & answer→MCQ 111 Mark
The number of words that can be formed out of the letters of the word $"\text{ARTICLE}"$ so that vowels occupy even places is :
AnswerThe word $\text{ARTICLE}$ consists of $3$ vowels that have to be arranged in the three even places.
This can be done in $3!$ ways.
And, the remaining $4$ consonants can be arranged among themselves in $4!$ ways.
$\therefore$ Total number of ways $= 3! \times 4! = 144$
View full question & answer→MCQ 121 Mark
The number of ways to arrange the letters of the word $\text{CHEESE}$ are :
AnswerTotal number of arrangements of the letters of the word $\text{CHEESE} =$ Number of arrangements of $6$ things taken all at a time, of which $3$ are of one kind $=\frac{6!}{3!}=120$
View full question & answer→MCQ 131 Mark
The number of different ways in which $8$ persons can stand in a row so that between two particular persons $A$ and $B$ there are always two persons, is:
- ✓
$60 \times 5!$
- B
$15 \times 4! \times 5!$
- C
$4! \times 5!$
- D
AnswerCorrect option: A. $60 \times 5!$
The four people, i.e $A, B$ and the two persons between them are always together. Thus, they can be considered as a single person.
So, along with the remaining $4$ persons, there are now total $5$ people who need to be arranged. This can be done in $5 !$ ways.
But, the two persons that have to be included between $A$ and $B$ could be selected out of the remaining $6$ people in ${ }^6 P _2$ ways, which is equal to $30$ .
For each selection, these two persons standing between $A$ and $B$ can be arranged among themselves in $2$ ways.
$\therefore$ Total number of arrangements $=5!\times 30 \times 2=60 \times 5!$
View full question & answer→MCQ 141 Mark
The product of $r$ consecutive positive integers is divisible by :
- ✓
$r!$
- B
$(r − 1)!$
- C
$(r + 1)!$
- D
AnswerThe product of r consecutive integers is equal to $r!,$ so it will be divisible by $r!.$
View full question & answer→MCQ 151 Mark
In a room there are $12$ bulbs of the same wattage, each having a separate switch. The number of ways to light the room with different amounts of illumination is:
- A
$12^2 − 1$
- B
$2^{12}$
- ✓
$2^{12} − 1$
- D
AnswerCorrect option: C. $2^{12} − 1$
Each of the bulb has its own switch, i.e each bulb will have two outcomes $−$ it will either glow or not glow. Thus, each of the $12$ bulbs will have $2$ outcomes.
$\therefore $ Total number of ways to illuminate the room $= 2^{12}$
Here, we have also considered the way in which all the bulbs are switched$-$off. However, this is not required as we need to find out only the number of ways of illuminating the room.
Hence, we subtract that one way from the total number of ways.
$= 2^{12}− 1$
View full question & answer→MCQ 161 Mark
The number of words that can be made by re $-$ arranging the letters of the word $\text{APURBA}$ so that vowels and consonants are alternate is :
AnswerThe word $\text{APURBA}$ is a $6$ letter word consisting of $3$ vowels that can be arranged in $3$ alternate places, in $\frac{3!}{2!}$ ways.
The remaining $3$ consonants can be arranged in the remaining $3$ places in $3!$ ways.
$\therefore$ Total number of words that can be formed $=\frac{3!}{2!}\times3!=18$
But this whole arrangement can be set $-$ up in total two ways,
i.e either $\text{VCVCVC}$ or $\text{CVCVCV}$.
$\therefore$ Total number of words $= 18 \times 2 = 36$
View full question & answer→MCQ 171 Mark
The number of arrangements of the word $"\text{DELHI}"$ in which $E$ precedes $I$ is :
AnswerThere are $4$ cases where $E$ precedes $I$ i.e.
Case $1$: When $E$ and $I$ are together, which are possible in $4$ ways whereas other $3$ letters are arranged in $3!,$
So, the number of arrangements $= 4 \times 3! = 24$
Case $2$ : When $E$ and $I$ have $1$ letter in between, which are possible in $3$ ways whereas other $3$ letters are arranged in $3!,$
So,the number of arrangements $= 3 \times 3! = 18$
Case $3$ : When $E$ and $I$ have $2$ letters in between, which are possible in $2$ ways whereas other $3$ letters are arranged in $3!,$
So,the number of arrangements $= 2 \times 3! = 12$
Case $4$ : When $E$ and $I$ have $3$ letters in between, which are possible in $1$ way whereas other $3$ letters are arranged in $3!,$
So,the number of arrangements $= 1 \times 3! = 6$
Thus, total number of arrangements $= 24 + 18 +12 + 6 = 60$
View full question & answer→MCQ 181 Mark
The number of arrangements of the letters of the word $\text{BHARAT}$ taking $3$ at a time is:
AnswerWhen we make words after selecting letters of the word $\text{BHARAT,}$ it could consist of a single $A,$ two As or no $A.$
Case$-I:$ $A$ is not selected for the three letter word.
Number of arrangements of three letters out of $B, H, R$ and $T = 4 \times 3 \times 24 \times 3 \times 2 = 24$
Case$-II:$ One $A$ is selected and the other two letters are selected out of $B, H, R$ or $T.$ Possible ways of selection: Selecting two letters out of $B, H, R$ or $T$ can be done in $^4P_2 = 12$ ways. Now, in each of these $12$ ways, these two letters can be placed at any of the three places in the three letter word in $3$ ways.
$\therefore$ Total number of words that can be formed $= 12 \times 3 = 36$
Case$-III:$ Two $A's$ and a letter from $B, H, R$ or $T$ are selected.
Possible ways of arrangement:
Number of ways of selecting a letter from $B, H, R$ or $T = 4$ And now this letter can be placed in any one of the three places in the three letter word other than the two $A's$ in $3$ ways.
$\therefore$ Total number of words having $2 A's = 4 \times 3 = 12$
Hence, total number of words that can be formed $= 24 + 36 + 12 = 72$
View full question & answer→MCQ 191 Mark
How many numbers greater than $10$ lacs be formed from $\{2, 3, 0, 3, 4, 2, 3\}$?
Answer$10$ lakhs consists of seven digits.
Number of arrangements of seven numbers of which $2$ are similar of first kind, $3$ are similar of second kind $=\frac{7!}{2!\ 3!}$
But, these numbers also include the numbers in which the first digit has been considered as $0$.
This will result in a number less than $10$ lakhs.
Thus, we need to subtract all those numbers.
Numbers in which the first digit is fixed as $0 =$ Number of arrangements of the remaining $6$ digits $=\frac{6!}{2!\ 3!}$
Total numbers greater than $10$ lakhs that can be formed using the given digits $=\frac{7!}{2!\ 3!}-\frac{6!}{2!\ 3!}$
$=420-60$
$=360$
$\text{x}\in(-\infty,-7)\cup(11,\infty)$
View full question & answer→MCQ 201 Mark
Number of all four digit numbers having different digits formed of the digits $\{1, 2, 3, 4\}$ and $5$ and divisible by $4$ is :
AnswerIn order to make a number divisible by $4,$ its last two digits must be divisible by $4,$ which in this case can be $\{ 12, 24, 32\}$ or $52$.
Since repetition of digits is not allowed, the remaining first two digits can be arranged in $3 \times 2$ ways in each case.
$\therefore$ Total number of numbers that can be formed $= 4 \times {3 \times 2} = 24$
View full question & answer→MCQ 211 Mark
The number of six letter words that can be formed using the letters of the word $"\text{ASSIST}"$ in which $S
\ ' s$ alternate with other letters is :
AnswerAll $S\ 's$ can be placed either at even places or at odd places,
i.e. in $2$ ways.
The remaining letters can be placed at the remaining places in $3!,$
i.e. in $6$ ways.
$\therefore$ Total number of ways $= 6 \times 2 \times 2 = 12$
View full question & answer→MCQ 221 Mark
The number of words from the letters of the word $'\text{BHARAT}\ '$ in which $B$ and $H$ will never come together, is :
AnswerTotal number of words that can be formed of the letters of the word $\text{BHARAT} =\frac{6!}{2!}$
Number of words in which the letters $B$ and $H$ are always together $=2\times\frac{5!}{2!}$
$=120$
$\therefore$ Number of words in which the letters $B$ and $H$ are never together $=360-120$
$=240$
View full question & answer→