MCQ 511 Mark
If $\ ^5\text{P}_\text{r}=\ ^{26}\text{p}_\text{r}-1, $ then the value of r is:
View full question & answer→MCQ 521 Mark
There are $5$ doors to a lecture hall.The number of ways that a student can enter the hall and leave it by a different door is:
AnswerA student has $5$ different doors to enter,
Every door a student enters through, there are $4$ more doors to leave through.
So,
Total no. of ways $=5\times4=20$
View full question & answer→MCQ 531 Mark
Amy and Adam are making boxes of truffles to give out as wedding favors. They have an unlimited supply of $5$ different types of truffles. If each box holds $2$ truffles of different types, how many different boxes can they make?
Answer$10$ boxes In every combination, $2$ types of truffles will be in the box, and $3$ types of truffles will not.
Therefore, this problem is a question about the number of anagrams that can be made from the "word" $\text{YYNNN:}$
$\frac{5!}{2!3!}=\frac{5\times4\times3\times2\times1}{3\times2\times1\times2\times1}$
$=5\times2$
$=10$
View full question & answer→MCQ 541 Mark
Permutation relates to the act of arranging all the members of a set into some sequence or order.
AnswerTrue, permutation relates to the act of arranging all the members of a set into some sequence or order.
View full question & answer→MCQ 551 Mark
A group consists of $4$ couples in which each of the $4$ persons have one wife each.In how many ways could they be arranged in a straight line such that the men and women occupy alternate positions:
- ✓
$1152$
- B
$1278$
- C
$1296$
- D
$1176$
AnswerCorrect option: A. $1152$
Required no of ways $= 2 \times 4! \times 4!$
$= 1152$
View full question & answer→MCQ 561 Mark
The total number of $9$ digit numbers of different digits is:
- A
$99!$
- B
$9!$
- C
$8 \times 9!$
- ✓
$9 \times 9!$
AnswerCorrect option: D. $9 \times 9!$
Given digit in the number $= 9$
$1^{st}$ place can be filled $= 9$ ways $= 9 ($from $1 - 9$ any number can be placed at first position$)$
$2^{nd}$ place can be filled $= 9$ ways $($from $0 - 9$ any number can be placed except the number which is placed at the first position$)$
$3^{rd}$ place can be filled $= 8$ ways.
$4^{th}$ place can be filled $= 7$ ways.
$5^{th}$ place can be filled $= 6$ ways.
$6^{th}$ place can be filled $= 5$ ways.
$7^{th}$ place can be filled $= 4$ ways.
$8^{th}$ place can be filled $= 3$ ways.
$9^{th}$ place can be filled $= 2$ ways.
So total number of ways $= 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2$
$= 9 \times 9!$
View full question & answer→MCQ 571 Mark
The number of ways in which $7$ pictures can be hung from $5$ picture nails on the wall is:
AnswerCorrect option: C. $2520$
View full question & answer→MCQ 581 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 $\text{A, I, K, N, R}$ and $\text{S.}$
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 $\text{KA} =$ Number of arrangements of the remaining $4$ letters = $4!$
The number of words beginning with $\text{KA} =$ Number of arrangements of the remaining $4$ letters $= 4!$
The number of words starting with $\text{KN} =$ Number of arrangements of the remaining $4$ letters $= 4!$
Alphabetically, the next letter will be $\text{KR.}$
Number of words starting with $\text{KR}$ 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 591 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 601 Mark
A garrison of $nn$ men had enough food to last for $30$ days. After $10$ days, $50$ more men joined them. If the food now lasted for $1616$ days, what is the value of $n?$
AnswerAfter $10$ days, the food for $n$ men is there for $20$ days.
This food can be eaten by $(\text{n}+50)$ men in $16$ days.
$\therefore20\text{n}=16(\text{n}+5)$
$\therefore\text{n}=200$
View full question & answer→MCQ 611 Mark
The number of ways in which $10$ different diamonds can be arranged to form a necklace, is:
- ✓
$181440$
- B
$161400$
- C
$261960$
- D
AnswerCorrect option: A. $181440$
View full question & answer→MCQ 621 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
$^{\text{n}}\text{P}_{\text{r}-3}$
- ✓
$\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
AnswerCorrect option: D. $\text{r! }^{ \text{n}-3}\text{C}_{\text{r}-3}$
Here, 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 631 Mark
$15$ buses operate between Hyderabad and Tirupathi.The number of ways can a man go to Tirupathi from Hyderabad by a bus and return by a different bus is:
AnswerWhile going ,number of ways to choose $1$ bus out of $15$ is $^{15}C_1$.
While return trip, in order to come with different bus, number of buses left $= 14.$
Number of ways to choose $1$ bus for return trip $=\ ^{14}C_1$
So, the required number of ways for going and return $=\ ^{15}C_1×^{14}C_1 = 210$
View full question & answer→MCQ 641 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 651 Mark
If ${^\text{20}}\text{C}_{\text{r}}={^\text{20}}\text{C}_{\text{r-10}}$ is then ${^\text{18}}\text{C}_{\text{r}}$ equal to:
Answer$\text{r}+\text{r}-10=20$
$\Rightarrow 2\text{r}-10=20$
$\Rightarrow 2\text{r}=30$
$\Rightarrow \text{r}=15$
Now,
${^\text{18}}\text{C}_{\text{r}}={^\text{18}}\text{C}_{\text{15}}$
$\therefore\ {^\text{18}}\text{C}_{\text{15}}={^\text{18}}\text{C}_{\text{3}}$
$\therefore\ {^\text{18}}\text{C}_{\text{3}}=\frac{18}{3}\times\frac{17}{2}\times16$
$=816$
View full question & answer→MCQ 661 Mark
A coin is tossed $n$ times, the number of all the possible outcomes is:
- A
$2n$
- ✓
$2^n$
- C
$C(n, 2)$
- D
$P(n, 2)$
AnswerWe know that, when a coin is tossed, we will get either head or tail.
Therefore, the number of all possible outcomes when a coin is tossed $n$ times is $2^n$.
View full question & answer→MCQ 671 Mark
The greatest number that can be formed by the digits $7, 0, 9, 8, 6, 3$
- A
$9, 87, 360$
- B
$9, 87, 063$
- ✓
$9, 87, 630$
- D
$9, 87, 603$
AnswerCorrect option: C. $9, 87, 630$
The greatest number that can be formed by the digits $7, 0, 9, 8, 6, 3$ is $9\ 8\ 7\ 6\ 3\ 0$ To achieve this arrange the given numbers in descending order.
View full question & answer→MCQ 681 Mark
Five persons $\text{A, B, C, D}$ and $\text{E}$ occupy seats in a row such that $\text{A}$ and $\text{B}$ sit next to each other.In how many possible ways can these five people sit:
Answer$4! \times 2$ ways.
i.e, $24 \times 2 = 48$ ways
View full question & answer→MCQ 691 Mark
Seven different lecturers are to deliver lectures in seven periods of a class on a particular day. $A B,$ and $C$ are three of the lecturers.The umber of ways in which a routine for the day can be made such that $A$ delivers his lecture before $B$ and $B$ before $C,$ is:
View full question & answer→MCQ 701 Mark
A batsman can score $0, 1, 2, 3, 4$ or $6$ runs from a ball. The number of different sequences in which he can score exactly $30$ runs in an over of six balls is:
View full question & answer→MCQ 711 Mark
Choose the correct answer. The number of triangles that are formed by choosing the vertices from a set of $12$ points, seven of which lie on the same line is.
AnswerTotal number of triangle formed from $12$ points taking $3$ at a time $=\ ^{12}C_3$
But given that out of $12$ points $7$ are collinear.
So, these seven points will form no triangle.
$\therefore$ The required number of triangles $=\ ^{12}C_3\ –\ ^{7}C_3$
$=\frac{12!}{3!\ 9!}-\frac{7!}{3!\ 4!}$
$=\frac{12\times11\times10\times9}{3\times2\times1\times9!}-\frac{7\times6\times5\times4!}{3\times2\times1\times4!}$
$=\frac{12\times11\times10}{3\times2}-\frac{7\times6\times5}{3\times2}$
$=220-35$
$=185$
View full question & answer→MCQ 721 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 731 Mark
The number of ways in which four particular persons $\text{A, B, C, D,}$ and six more persons can stand in a queue so that $A$ always stands before $\text{B B,}$ before $\text{C}$ and $\text{C}$ before $\text{D,}$ is:
- A
$7! 4!$
- B
$10!-7! 4! $
- ✓
$\frac{10!}{4!}$
- D
$\text{None of these}$
AnswerCorrect option: C. $\frac{10!}{4!}$
View full question & answer→MCQ 741 Mark
Find the number of permutations if $n = 12$ and $r = 2.$
AnswerThe solution is here:
$\text{n}=12$
$\text{r}=2$
Using the formula given above:
Permutation:
$\ ^\text{n}\text{p}_\text{r}$
$=\frac{\text{(n})!}{\text{(n-r)}!}$
$=\frac{(12)!}{(12-2)}$
$=\frac{12!}{10!}$
$=\frac{(12\times11\times10!)}{10!}$
$=132$
View full question & answer→MCQ 751 Mark
If $^{(\text{a}^2-\text{a})}\text{C}_{\text{2}}=^{(\text{a}^2-\text{a})\text{}}\text{C}_{\text{4}},$ is then $x:$
Answer$\text{a}^{2}-\text{a}=2+4$
$\Rightarrow \text{a}^{2}-\text{a}-6=0$
$\Rightarrow \text{a}^{2}-3\text{a}+2\text{a}-6=0$
$\Rightarrow \text{a}(\text{a}-3)+2(\text{a}-3)=0$
$\Rightarrow (\text{a}+2)(\text{a}-3)=0$
$\Rightarrow \text{a}=-2,\text{a}=3$
$\text{a}=3$
View full question & answer→MCQ 761 Mark
Total number of divisors of $5880$ is equal to:
View full question & answer→MCQ 771 Mark
The exponent of $3$ in $100!$ is:
View full question & answer→MCQ 781 Mark
How many factors are $2^5\times 3^6 \times 5^2$ are perfect squares:
AnswerAny factors of $2^5\times 3^6 \times 5^2$ which is a perfect square will be of the form $2^a \times 3^b \times 5^c$
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 $= 3 \times 4 \times 2 = 24$
View full question & answer→MCQ 791 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 $^6P_2$ 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 801 Mark
There are $6$ letters and $6$ directed envelopes.Find the number of ways in which all letters are put in the wrong envelopes.
View full question & answer→MCQ 811 Mark
The number of ways in which $6$ men add $5$ women can dine at a round table, if no two women are to sit together, is given by:
- A
$30$
- ✓
$5! \times 5!$
- C
$5! \times 4!$
- D
$7! \times 5!$
AnswerCorrect option: B. $5! \times 5!$
Again, $6$ girls can be arranged among themselves in $5!$ ways in a circle.
So, the number of arrangements where boys and girls sit attentively in a circle $= 5! \times 5!$
View full question & answer→MCQ 821 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 831 Mark
Choose the correct answer.The number of words which can be formed out of the letters of the word $\text{ARTICLE,}$ so that vowels occupy the even place is.
- A
$1440$
- ✓
$144$
- C
$7!$
- D
$^4C_4 \times ^3C_3$
AnswerTotal number of letters in the$ ‘\text{ARTICLE’}$ is $7$ out which $\text{A, E, I}$ are vowels and $\text{R, T, C, L}$ are consonants
Given that vowels occupy even place
$\therefore$ Possible arrangement can be shown as below
$\text{C, V, C, V, C, V, C}$
i.e. on $2^{\text {nd }}, 4^{\text {th}}$ and $6^{\text {th }}$ places
Therefore, number of arrangement $={ }^3 P_3=3!=6$ ways
Now consonants can be placed at $1,3,5$ and $7^{\text {th }}$ place
$\therefore$ Number of arrangement $={ }^4{P}_4=4!=24$
So, the total number of arrangements $= 6 \times 24 = 144$
View full question & answer→MCQ 841 Mark
The value of $P(n, n - 1)$ is:
AnswerWe know that $\text{P}(\text{n}, \text{r}) = \ ^\text{n}\text{P}_\text{r} = \frac{\text{n}!}{(\text{n}-\text{r})!}$
Hence, $\text{P}(\text{n}, \text{r}) = \ ^\text{n}\text{P}_\text{n-1} = \frac{\text{n}!}{\big[\text{n}-\text{r}\big]!}$
$\text{P}(\text{n, n-1}) = \frac{\text{n!}}{(\text{n}-\text{n}+1)!} = \frac{\text{n}!}{1!} = \text{n}!$
Therefore, the value of $\text{P}(\text{n}, \text{n}-1)$ is $\text{n}!.$
View full question & answer→MCQ 851 Mark
The number of ways $4$ boys and $3$ girls can be seated in a row so that they are alternate is:
AnswerGiven that, there are $4$ boys and $3$ girls.
The only pattern $4$ boys and $3$ girls are arranged in an alternate way is $\text{BGBGBGB.}$
Therefore, the total number of ways is $4! \times 3! = 144.$
View full question & answer→MCQ 861 Mark
If $={^\text{43}}\text{C}_{\text{r-6}}={^\text{43}}\text{C}_{\text{3r+1}},$ then the value of $r$ is is:
Answer$\text{r}-6+3\text{r}+1=43$
$\Rightarrow 4\text{r}-5=43$
$\Rightarrow 4\text{r}=48$
$\Rightarrow \text{r}=12$
View full question & answer→MCQ 871 Mark
$6$ men and $4$ women are to be seated in a row so that no two women sit together.The number of ways they can be seated is:
- ✓
$604800$
- B
$17280$
- C
$120960$
- D
$518400$
AnswerCorrect option: A. $604800$
$6$ men can be sit as $\text{M}\times\text{M}\times\text{M}\times\text{M}\times\text{M}\times\text{M}$
Now, there are $7$ spaces and $4$ women can be sit as $ \ ^7\text{P}_4 = \ ^7\text{P}_3 = \frac{7!}{3!} = \frac{(7\times6\times5\times4\times3!)}{3!}$
$=7\times6\times5\times4$
$=840$
Now, total number of arrangement $= 6!\times840$
$= 720\times840$
$= 604800$
View full question & answer→MCQ 881 Mark
How many different signals can be transmitted by arranging $3$ red, $2$ yellow and $2$ green flags on a pole: $[$Assume that all the $7$ flags are used to transmit a signal$].$
AnswerHere, $\text{n}=3+2+2=7$
$\text{P}_1=3,\text{P}_2=2$ and $\text{P}_3=2$
$\therefore$ Required number of different signals
$=\frac{\text{n}!}{\text{p}_1!\text{p}_1\text{p}_1}$
$=\frac{7!}{3!2!1!}$
$=\frac{7.6.5.4}{2.2}$
$=7.6.5$
$=210$
View full question & answer→MCQ 891 Mark
On the eve of Diwali festival, a group of $12$ friends greeted every other friend by sending greeting cards. Find the number of cards purchased by the group:
AnswerThere being $12$ friends in the group, each friend must have purchased $(12 - 1)$
i.e. $11$ cards for sending greeting to rest of his $11$ friends.
Thus total number of cards purchased by all the friends together is $12 \times 11$
i.e. $132.$
View full question & answer→MCQ 901 Mark
The number of ways in which $8$ students can be seated in a line is:
- A
$5040$
- B
$50400$
- C
$40230$
- ✓
$40320$
AnswerCorrect option: D. $40320$
For the $1^{st}$ position, there are $8$ possible choices.
For the $2^{nd}$ position, there are $7$ possible choices.
For the $3^{rd}$ position, there are $7$ possible choices, etc.
And for the eighth position, there is only one possible choice.
Hence, this can be written as $8!$
$($i.e.$) 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 40,320$
View full question & answer→MCQ 911 Mark
How many numbers amongst the numbers $9$ to $54$ are there which are exactly divisible by $9$ but not by $3?$
AnswerAny number divisible by $9$ is also divisible by $3.$
View full question & answer→MCQ 921 Mark
The number of products that can be formed with $10$ prime numbers taken two or more at a time is:
- A
$2^{10}$
- B
$2^{10}- 1$
- ✓
$2^{10}- 11$
- D
$2^{10}-10$
AnswerCorrect option: C. $2^{10}- 11$
View full question & answer→MCQ 931 Mark
Choose the correct answer. If $^nC_{12}$=$^nC_8$ , then n is equal to.
AnswerGive that $^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_8[\because\ ^\text{n}\text{C}_\text{r}=\ ^\text{n}\text{C}_\text{n-r}]$
$^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_\text{n-8}$
$\therefore\text{n}-8=12$
$\Rightarrow\text{n}=12+8=20$
View full question & answer→MCQ 941 Mark
Each combination corresponds to many permutations:
AnswerIn combination.
Each combination can be considered as a set of selection an order.
Each selection has a defined order.
They can be considered as a permutation.
Each cpmbination corresponds to many permutations.
Hence the above statement is true.
View full question & answer→MCQ 951 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 961 Mark
A lady gives a dinner party for six guests. The number of ways in which they may be selected from among ten friends if two of the friends will not attend the party together is:
AnswerSuppose there are two friends, $A$ and $B,$ who do not attend the party together.
If both of them do not attend the party, then the number of ways of selecting $6$ guests $={^\text{8}}\text{C}_{\text{6}}=28$
If one of them attends the party, then the number of ways of selecting $6$ guests $=2{^\text{8}}\text{C}_{\text{5}}=112$
Total number of ways $= 112 + 28 = 140$
View full question & answer→MCQ 971 Mark
A committee of $7$ has to be formed from $9$ boys and $4$ girls.In how many ways can this be done when the committee consists of at least $3$ girls.
AnswerGiven number of boys $= 9$
Number of girls $= 4$
Now, A committee of $7$ has to be formed from $9$ boys and $4$ girls.
Now, the committee consists of atleast $3$ girls:
$^4\text{C}_3\times\ ^9\text{C}_4+\ ^4\text{C}_4\times\ ^9\text{C}_3$
$= \bigg[\frac{4!}{(3! \times 1!)}\times\frac{9!}{(4! \times 5!)}\bigg] +\ ^9\text{C}_3$
$=[\frac{(4 \times 3!)}{3!}\times\frac{(9\times8\times7\times6\times5!)}{(4!\times5!)]} +\frac{9!}{(3!\times6!)}$
$= \bigg[\frac{4\times(9\times8\times7\times6)}{4!}\bigg] + \frac{(9\times8\times7\times6!)}{(3!\times6!)}$
$= \bigg[\frac{4\times(9\times8\times7\times6)}{4\times3\times2\times1}\bigg] + \frac{(9\times8\times7)}{(3!)}$
$=(9\times8\times7) + \frac{(9\times8\times7)}{(3\times2\times1)}$
$=504+\big(\frac{504}{6}\big)$
$=504+84$
$=588$
View full question & answer→MCQ 981 Mark
Two persons entered a Railway compartment in which $7$ seats were vacant.The number of ways in which they can be seated is:
Answer$\rightarrow$ The $1^{st}$ person can take one of the $7$ seats
$\rightarrow 2^{nd}$ person can take any one of the remaining $6$ seats.
$\Rightarrow$ So, the total $= 7 \times 6 = 42$
View full question & answer→MCQ 991 Mark
In a colony, there are $55$ members. Every member posts a greeting card to all the members. How many greeting cards were posted by them?
- A
$990$
- ✓
$2970$
- C
$1980$
- D
$890$
AnswerCorrect option: B. $2970$
First player can post greeting cards to the remaining $54$ players in $54$ ways.
Second player can post greeting card to the $54$ players.
Similarly, it happens with the rest of the players.The total numbers of greeting cards posted are.
$54 + 54 + 54 …$
$54 (55 \text{times}) = 54 \times 55 = 2970.$
View full question & answer→MCQ 1001 Mark
The value of $2 \times P(n, n - 2)$ is:
AnswerGiven, $ 2\times\text{P}(\text{n}, \text{n} – 2)$
$= 2\times\frac{\text{n!}}{(\text{n} – (\text{n} – 2))}$
$= 2\times\frac{\text{n!}}{(\text{n} – (\text{n}+2))}$
$= 2\times\big(\frac{\text{n!}}{2}\big)$
$= \text{n!}$
So, $ 2\times\text{P}(\text{n}, \text{n} – 2)=\text{n}!$
View full question & answer→