If the letter of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
A.
346
B.
140
C.
196
D.
280
Correct Answer: C
Explanation:
Case 1 :
No of ways student can answer 10 questions = ${}^5{C_4} \times {}^8{C_6}$ = 140
Case 2 :
No of ways student can answer 10 questions = ${}^5{C_5} \times {}^8{C_5}$ = 56
If ${}^n{C_r}$ denotes the number of combination of n things taken r at a time, then the expression $\,{}^n{C_{r + 1}} + {}^n{C_{r - 1}} + 2\, \times \,{}^n{C_r}$ equals
Five digit number divisible by 3 is formed using 0, 1, 2, 3, 4 and 5 without repetition. Total number of such numbers are :
A.
312
B.
3125
C.
120
D.
216
Correct Answer: D
Explanation:
Note : For a number to be divisible by 3, the sum of digits should be divisible by 3.
Here given numbers are 0, 1, 2, 3, 4 and 5. Out of those 6 numbers possible sets of 5 numbers are (1, 2, 3, 4, 5) and (0, 1, 2, 4, 5) whose sum are divisible by 3.
Set 1 : Set is = (1, 2, 3, 4, 5). Sum of digits = 1 + 2 + 3 + 4 + 5 = 15 (Divisible by 3)
So total no of arrangement = 1$ \times $2$ \times $3$ \times $4$ \times $5 = 5!
Set 2 : Set is = (0, 1, 2, 4, 5). Sum of digits = 0 + 1 + 2 + 4 + 5 = 12 (Divisible by 3)
So total no of arrangement = 4$ \times $4$ \times $3$ \times $2$ \times $1 = 4.4!
Number greater than 1000 but less than 4000 is formed using the digits 0, 1, 2, 3, 4 (repetition allowed). Their number is :
A.
125
B.
105
C.
374
D.
625
Correct Answer: C
Explanation:
There are 3 possible ways that we can make number greater than 1000 but less than 4000 using the digits 0, 1, 2, 3, 4 where repetition is allowed
Case 1 : First digit is 1 = 1 _ _ _
Possible numbers starting with 1 = 1$ \times $5$ \times $5$ \times $5 = 125
But this includes 1000 also which does not satisfy the given condition of being greater than 1000. Hence there will be 124 numbers having 1 in the first place.
Case 2 : First digit is 2 = 2 _ _ _
Possible numbers starting with 2 = 1$ \times $5$ \times $5$ \times $5 = 125
Case 3 : First digit is 3 = 3 _ _ _
Possible numbers starting with 3 = 1$ \times $5$ \times $5$ \times $5 = 125
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is
Consider 4 boxes, where each box contains 3 red balls and 2 blue balls. Assume that all 20 balls are distinct. In how many different ways can 10 balls be chosen from these 4 boxes so that from each box at least one red ball and one blue ball are chosen ?
A.
21816
B.
85536
C.
12096
D.
156816
Correct Answer: A
Explanation:
Case 1 :
Among four bags from one bag 4 balls are taken. Number of ways of chosing one bag from 4 bags = ${}^4{C_1}$
From this bag number of ways of taking 4 balls are
(a) 3 Red and 1 Blue balls, which can be chosen in ${}^3{C_3} \times {}^2{C_1}$ ways.
(b) 2 Red and 2 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_2}$ ways.
$ \therefore $ Total number of ways of choosing this 4 balls from this bag = ${}^4{C_1}$$\left( {{}^3{C_3} \times {}^2{C_1} + {}^3{C_2} \times {}^2{C_2}} \right)$
Now, two balls are taken from remaining three bags.
From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.
So, for three bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^3}$ ways.
$ \therefore $ Total number of ways of chosing 10 balls from these four bags
Among four bags from two bags 3 balls are taken. Number of ways of chosing two bags from 4 bags = ${}^4{C_2}$
From each bag, number of ways of taking 3 balls are
(a) 2 Red and 1 Blue balls, which can be chosen in ${}^3{C_2} \times {}^2{C_1}$ ways.
(b) 1 Red and 2 Blue balls, which can be chosen in ${}^3{C_1} \times {}^2{C_2}$ ways.
So, for two bags three balls can be taken in ${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$ ways.
$ \therefore $ Total number of ways of choosing this 3 balls from this two bags = ${}^4{C_2}$${\left( {{}^3{C_2} \times {}^2{C_1} + {}^3{C_1} \times {}^2{C_2}} \right)^2}$
Now, two balls are taken from remaining two bags.
From each bag two balls can be taken in ${{}^3{C_1} \times {}^2{C_1}}$ ways.
So, for two bags two balls can be taken in ${\left( {{}^3{C_1} \times {}^2{C_1}} \right)^2}$ ways.
$ \therefore $ Total number of ways of chosing 10 balls from these four bags
From Case 1 and Case 2, total number of ways of chosing 10 balls from these 4 boxes so that from each box at least one red ball and one blue ball are chosen
In a high school, a committee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5.
(i) Let $\alpha $1 be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls.
(ii) Let $\alpha $2 be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.
i) Let $\alpha $3 be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls.
(iv) Let $\alpha $4 be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both M1 and G1 are NOT in the committee together.
LIST-I
LIST-II
P. The value of $\alpha_1$ is
1. 136
Q. The value of $\alpha_2$ is
2. 189
R. The value of $\alpha_3$ is
3. 192
S. The value of $\alpha_4$ is
4. 200
5. 381
6. 461
The correct option is
A.
P $ \to $ 4; Q $ \to $ 6; R $ \to $ 2; S $ \to $ 1
B.
P $ \to $ 1; Q $ \to $ 4; R $ \to $ 2; S $ \to $ 3
C.
P $ \to $ 4; Q $ \to $ 6; R $ \to $ 5; S $ \to $ 2
D.
P $ \to $ 4; Q $ \to $ 2; R $ \to $ 3; S $ \to $ 1
Correct Answer: C
Explanation:
Given 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5
(i) $\alpha $1 $ \to $ Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.
(ii) $\alpha $2 $ \to $ Total number of ways selecting at least 2 member and having equal number of boys and girls i.e., ${}^6{C_1}{}^5{C_1} + {}^6{C_2}{}^5{C_2} + {}^6{C_3}{}^5{C_3} + {}^6{C_4}{}^5{C_4} + {}^6{C_5}{}^5{C_5}$
= 30 + 150 + 200 + 75 + 6 = 461
$\alpha $2 = 461
(iii) $\alpha $3 $ \to $ Total number of ways of selecting 5 members in which at least 2 of them girls
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be select from this club including the selection of a captain (from among these 4 members ) for the team. If the team has to include at most one boy, then the number of ways of selecting the team is
A.
380
B.
320
C.
260
D.
95
Correct Answer: A
Explanation:
We have 6 girls and 4 boys in the club and we have to select team of 4 members in which one is captain and at most one boy.
$\therefore$ We can have one boy and three girls in team or all four girls.
So, selection of 1 boy from 4 boys $={ }^4 C_1$ ways
Selection of 3 girls from 6 girls $={ }^6 \mathrm{C}_3$ ways
Selection of 4 girls from 6 girls $={ }^6 \mathrm{C}_4$ ways
$\therefore$ Total number of ways selecting the team
$
=\left({ }^4 \mathrm{C}_1 \cdot{ }^6 \mathrm{C}_3+{ }^6 \mathrm{C}_4\right) \times 4
$
(since, among any of the selection we have 4 choice to select captain)
$
\begin{aligned}
\therefore \text { Total number of ways } & =(4 \times 20+15) \times 4 \\\\
& =380
\end{aligned}
$
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is
A.
264
B.
265
C.
53
D.
67
Correct Answer: C
Explanation:
Given, six cards and six envelops are numbered $1,2,3,4,5$ and 6.
In the above dearrangement, there are 5 ways in which card number 1 is going wrong envelope i.e, other than envelope number 1. So, when card number 1 going in envelop number 2 is $\frac{265}{5}=53$ ways.
Hint:
In the total dearrangement of 6 cards there are 5 ways in which card number 1 is going other than envelope number 1.
Let ${{a_n}}$ denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0.Let ${{b_n}}$ = the number of such n-digit integers ending with digit 1 and ${{c_n}}$ =the number of such n-digit integers ending with digit 0.
The value of ${{b_6}}$ is
A.
7
B.
8
C.
9
D.
11
Correct Answer: B
Explanation:
Given, $b_n$ denotes the number of $n$-digit integer formed by the digits 0, 1 or both such that $n$-digit integer ending with 1 and no consecutive digits are '0'.
$\therefore \quad b_6=$ six digit number ending with 1.
Like 1 ........... 1, and rest four places are filled as Case No. (I) : Use four ' 1 '
$\text { Case No. (I) : Use four ' } 1 \text { ' }$
Let ${{a_n}}$ denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0.Let ${{b_n}}$ = the number of such n-digit integers ending with digit 1 and ${{c_n}}$ =the number of such n-digit integers ending with digit 0.
Which of the following is correct?
A.
${a_{17}} = {a_{16}} + {a_{15}}$
B.
${c_{17}} \ne {c_{16}} + {c_{15}}$
C.
${b_{17}} \ne {b_{16}} + {c_{16}}$
D.
${a_{17}} = {c_{17}} + {b_{16}}$
Correct Answer: A
Explanation:
For $a_n$
Case I : If the unit digit is 1, and rest $(n-1)$ places are filled as $a_{n-1}$
The total number of ways in which 5 balls of different colours can be distributed among 3 persons so that each person gets at least one ball is
A.
75
B.
150
C.
210
D.
243
Correct Answer: B
Explanation:
Here, 5 distinct balls are to be distributed amongst 3 persons so that each gets at least one
ball. So, two possible cases arises
Case I : Two of the persons get one-one ball each and the third person gets three balls.
i.e.
A
B
C
1
1
3
Now, A can get the ball in ${ }^5 \mathrm{C}_1$ ways. After that, $B$ can get one ball in ${ }^4 C_1$ ways and then after $C$ can get three balls in ${ }^3 \mathrm{C}_3$ ways.
Case II : Two of the persons get two-two balls each and the third person gets one ball.
i.e.
A
B
C
2
2
1
Now, A can get two balls in ${ }^5 \mathrm{C}_2$ ways. After that, $B$ can get 2 ball in ${ }^3 \mathrm{C}_2$ ways and then after $C$ can get 1 ball in ${ }^1 C_1$ way. Hence, total number of ways
Consider all possible permutations of the letters of the word ENDEANOEL. Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.
Column I
Column II
(A)
The number of permutations containing the word ENDEA is
(P)
5!
(B)
The number of permutations in which the letter E occurs in the first and the last position is
(Q)
2 $\times$ 5!
(C)
The number of permutations in which none of the letters D, L, N occurs in the last five positions is
(R)
7 $\times$ 5!
(D)
The number of permutations in which the letters A, E, O occur only in odd positions is
(S)
21 $\times$ 5!
A.
(A) - p ; (B) - s; (C) - q ; (D) - q
B.
(A) - q ; (B) - q ; (C) - s ; (D) - p
C.
(A) - p ; (B) - s; (C) - p ; (D) - r
D.
(A) - p ; (B) - r ; (C) - q ; (D) - p
Correct Answer: A
Explanation:
(A) Considering ENDEA as one group, remaining letters are N, O, E, L.
So, no. of permutations = 5!
(A) - (i)
(B) E occurs in 1st and last positions. The remaining letters are N, N, D, A, O, E, L.
The letters of the word COCHIN are permuted and all the permutations are arranged in an alphabetical order as in an English dictionary. The number of words that appear before the word COCHIN is :
A.
360
B.
192
C.
96
D.
48
Correct Answer: C
Explanation:
The letter of word COCHIN in alphabetic order are C, C, H, I, N, O
Fixing $1^{\text {st }}$ letter $\mathrm{C}$ and keeping $\mathrm{C}$ at second place, rest 4 can be arranged in 4 ! ways.
Similarly, the words starting with $\mathrm{CH}, \mathrm{CI}, \mathrm{CN}$ are 4 ! in each case
Then fixing first two letters as CO next four places, when filled in alphabetic order give the word COCHIN.
$\therefore$ Number of words coming before COCHIN are $4 \times 4$!
A rectangle with sides of lenght (2m - 1) and (2n - 1) units is divided into squares of unit lenght by drawing parallel lines as shown in the diagram, then the number of rectangles possible with odd side lengths is
Let ${T_n}$ denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If ${T_{n + 1}} - {T_n} = 21$, then n equals
A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5, without repetition. The total number of ways this can be done is
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4; and then the men select the chairs from amongst the remaining. The number of possible arrangements is
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the numbers of words which have at least one letter repeated are
Let the set of all relations $R$ on the set $\{a, b, c, d, e, f\}$, such that $R$ is reflexive and symmetric, and $R$ contains exactly $10$ elements, be denoted by $\mathcal{S}$.
Then the number of elements in $\mathcal{S}$ is ________________.
Correct Answer: 105
Explanation:
Problem restated: Let $R$ be a relation on $X = \{a,b,c,d,e,f\}$.
We want $R$ to be reflexive and symmetric, and contain exactly 10 elements.
Let $\mathcal{S} = \{R : R \text{ satisfies this}\}$. Find $|\mathcal{S}|$.
Step 1. Size of the universe
$|X| = 6$.
So $X \times X$ has $36$ ordered pairs.
Step 2. Reflexivity
Reflexive means we must include all $(x,x)$ pairs.
So there are 6 forced elements.
So any reflexive relation on $X$ automatically contains:
$ \{(a,a),(b,b),(c,c),(d,d),(e,e),(f,f)\}. $
So our reflexive relation starts with 6 elements.
Step 3. Symmetry
For off-diagonal elements:
If we include $(x,y)$ (for $x \neq y$), we must also include $(y,x)$.
So these off-diagonal pairs come in symmetric pairs (each of size 2).
There are $\binom{6}{2} = 15$ distinct unordered pairs $\{x,y\}$, each would correspond to either including both $(x,y)$ and $(y,x)$, or excluding both. So think of them as independent choices.
Step 4. Required size
Total size should be 10.
We already have 6 diagonal pairs.
So we need 4 more pairs.
But each symmetric pair contributes 2 elements.
So to get 4 additional, we must include exactly 2 symmetric pairs.
Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example, $2210222$ is in $S$, but $0210222$ is NOT in $S$.
Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$, is equal to ____________.
If the value of $n(Y)+n(Z)$ is $k^2$, then $|k|$ is _________.
Correct Answer: 36
Explanation:
given $|a-b| \geq 2$ so if
i.e. Total elements in X is ${ }^{20} \mathrm{C}_6$
Now for $\mathrm{n}(\mathrm{Y})$, range of R has exactly one element i.e. second elements must be constant in R and since R must have 6 element so it is not possible to satisfy both condition so $\mathrm{n}(\mathrm{Y})=0$.
A group of 9 students, $s_1, s_2, \ldots, s_9$, is to be divided to form three teams $X, Y$, and $Z$ of sizes 2,3 , and 4 , respectively. Suppose that $s_1$ cannot be selected for the team $X$, and $s_2$ cannot be selected for the team $Y$. Then the number of ways to form such teams, is ____________.
Correct Answer: 665
Explanation:
$\matrix{
x & y & z \cr
2 & 3 & 4 \cr
{{{\overline S }_1}} & {{{\overline S }_2}} & {} \cr
}$
C-i) When x does not contain S$_1$, but contains S$_2$
An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during 1-15 June 2021 is ...........
Correct Answer: 495
Explanation:
Let the engineer visits the factory first time after x1 days to 1 June, second time after x2 days to first visit and so on.
$ \therefore $ x1 + x2 + x3 + x4 + x5 = 11
where x1, x5 $ \ge $ 0 and x2, x3, x4 $ \ge $ 1 according to the requirement of the question.
Now, let x2 = a + 1, x3 = b + 1 and x4 = c + 1 where a, b, c $ \ge $ 0
$ \therefore $ New equation will be
x1 + a + b + c + x5 = 8
Now, the number of all possible ways in which the engineer can made visits is equals to the non-negative integral solution of equation
In a hotel, four rooms are available. Six persons are to be accommodated in these four rooms in such a way that each of these rooms contains at least one person and at most two persons. Then the number of all possible ways in which this can be done is ..........
Correct Answer: 1080
Explanation:
The groups of persons can be made only in 2, 2, 1, 1
$ \therefore $ So the number of required ways is equal to number of ways to distribute the 6 distinct objects in group sizes 1, 1, 2 and 2
Let |X| denote the number of elements in a set X. Let S = {1, 2, 3, 4, 5, 6} be a sample space, where each element is equally likely to occur. If A and B are independent events associated with S, then the number of ordered pairs (A, B) such that 1 $ \le $ |B| < |A|, equals .............
Correct Answer: 1523
Explanation:
Given sample space S = {1, 2, 3, 4, 5, 6} and let there are i elements in set A and j elements in set B.
Now, according to information 1 $ \le $ j < i $ \le $ 6.
When number of element in set B = 1 then number of elements in set A can be 2 or 3 or 4 or 5 or 6. Number of such pairs of A and B in this case
= 6C1[6C2 + 6C3 + 6C4 + 6C5 + 6C6]
When number of element in set B = 2 then number of elements in set A can be 3 or 4 or 5 or 6. Number of such pairs of A and B in this case
= 6C2[ 6C3 + 6C4 + 6C5 + 6C6]
When number of element in set B = 3 then number of elements in set A can be 4 or 5 or 6. Number of such pairs of A and B in this case
= 6C3[ 6C4 + 6C5 + 6C6]
When number of element in set B = 4 then number of elements in set A can be 5 or 6. Number of such pairs of A and B in this case
= 6C4[ 6C5 + 6C6]
When number of element in set B = 5 then number of elements in set A can be 6. Number of such pairs of A and B in this case
Five persons A, B, C, D and E are seated in a circular arrangement. If each of them is given a hat of one of the three colours red, blue and green, then the number of ways of distributing the hats such that the persons seated in adjacent seats get different coloured hats is ............
Correct Answer: 30
Explanation:
Given that no two persons sitting adjacent have hats of
same colour. Also, hats of different colour cannot be
used in 1 + 1 + 3 combination because any three hats
cannot be of same colour.
So, only possible combination due to circular arrangement is 2 + 2 + 1.
So, there are following three cases of selecting hats are
2R + 2B + 1G or 2B + 2G + 1R or 2G + 2R + 1B.
To distribute these 5 hats first we will select a person which we can done in ${{}^5{C_1}}$ ways and distribute that hat which is one of it's colour. And, now the remaining four hats can be distributed in two ways. So, total ways will be 3 $ \times $ ${{}^5{C_1}}$ $ \times $ 2 = 3 $ \times $ 5 $ \times $ 2 = 30
The number of 5 digit numbers which are divisible by 4, with digits from the set {1, 2, 3, 4, 5} and the repetition of digits is allowed, is .................
Correct Answer: 625
Explanation:
A number is divisible by 4 if last 2 digit number is divisible by 4.
$ \therefore $ Last two digit number divisible by 4 from (1, 2, 3, 4, 5) are 12, 24, 32, 44, 52
$ \therefore $ The number of 5 digit number which are divisible by 4, from the digit (1, 2, 3, 4, 5) and digit is repeated is 5 $ \times $ 5 $ \times $ 5 $ \times $ (5 $ \times $ 1) = 625
Words of length 10 are formed using the letters A, B, C, D, E, F, G, H, I, J. Let x be the number of such words where no letter is repeated; and let y be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then, ${y \over {9x}}$ = ?
Correct Answer: 5
Explanation:
The given, formed word is of length 10.
It is given that x is the number of words where no letter is repeated.
Also, it is given that y is the number of words where exactly one letter is repeated twice and no other letter is repeated. Therefore,
x = 10!
and y = 10C1 $\times$ 10C2 $\times$ 9C8 $\times$ 8!
Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of ${m \over n}$ is
Correct Answer: 5
Explanation:
Given: 5 boys and 5 girls
$n=$ number of ways of arranging them in a queue such that all the girls stand consecutively.
Let us consider 5 girls as one set
So, we have to arrange 5 boys and one set of girls. They can be arranged in 6 ! ways.
Also, the girls in the set can be arranged in 5 ! ways
So, total number of ways $=6 ! \times 5!$
$
\Rightarrow n=6 ! \times 5 !
$
Now, $m=$ number of ways of arranging them in a queue, such that exactly four girls stand consecutively.
$\because$ Exactly four girls can stand together so the remaining one girl must not stand consecutively with four girls.
Let us consider 2 cases:
Case I : The set of four girls is at the corner. Firstly, four girls are selected out of five girls in ${ }^5 \mathrm{C}_4$ ways. These girls are arranged in 4 ! ways.
Also, these girls can be placed in any of the two corners and the remaining one girl cannot stand next to the set of girls placed at the corner. So, the $5^{\text {th }}$ girl can stand at (7-1-1 = 5 ways) And the boys can be arranged in 5 ! ways.
Case II: The set of four girls are not placed at the corner.
So, four girls can be selected and arranged among themselves in ${ }^5 C_4 \times 4 !=5$ ! ways
These girls are not at the corner so they can be arranged at 5 places.
The $5^{\text {th }}$ girl can stand at $7-2-1=4$ ways. $\{$ As she cannot stand at places near the set of four girls $\}$ and the boys can be arranged in 5 ! ways.
So, number of ways $=5 ! \times 5 \times 5 \times 4 \times 5!$
Let ${n_1}\, < {n_2}\, < \,{n_3}\, < \,{n_4}\, < {n_5}$ be positive integers such that ${n_1}\, + {n_2}\, + \,{n_3}\, + \,{n_4}\, + {n_5}$ = 20. Then the number of such destinct arrangements $\,({n_1}\,,\,{n_2},\,\,{n_3},\,\,{n_4}\,,{n_5})$ is
Let ${n \ge 2}$ be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is
Correct Answer: 5
Explanation:
Number of blue lines $=n=$ number of sides of polygon so formed.
Number of red lines $={ }^n C_2-n$.
Thus, by joining $n$ points (not more than 2 on a line) there are ${ }^n C_2$ lines formed because for each line two points are required.
Also, red lines come after excluding sides of polygon.
Consider the set of eight vectors $V = \left\{ {a\,\hat i + b\,\hat j + c\hat k:a,\,b,\,c\, \in \left\{ { - 1,\,1} \right\}} \right\}$. Three non-coplanar vectors can be chosen from v in ${2^p}$ ways. Then p is
Correct Answer: 5
Explanation:
Given, the set of eight vectors
$\mathrm{V}=\{a \hat{i}+b \hat{j}+c \hat{k}: a, b, c \in\{-1,1\}\} .$
Now, the eight vectors are $\hat{i}+\hat{j}+\hat{k}, \hat{i}+\hat{j}-\hat{k}$,
Here, $\hat{i}+\hat{j}+\hat{k}$ and $-\hat{i}-\hat{j}-\hat{k}, \hat{i}+\hat{j}-\hat{k}$ and $-\hat{i}-\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+\hat{j}-\hat{k},-\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ are collinear vectors.
For the set of three non - coplanar vector, we have to select three set out of $\mathrm{S}_1, \mathrm{~S}_2, \mathrm{~S}_3, \mathrm{~S}_4$ and select one vector in every selected set of $\mathrm{S_1, S_2}, \mathrm{S}_3, \mathrm{~S}_4$
Recall that $\hat{i}+\hat{j}+\hat{k}$ and $-\hat{i}-\hat{j}-\hat{k}, \hat{i}+\hat{j}-\hat{k}$ and $-\hat{i}-\hat{j}+\hat{k}, \hat{i}-\hat{j}+\hat{k}$ and $-\hat{i}+\hat{j}-\hat{k},-\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}-\hat{k}$ are collinear vectors.
Let $\left( {x,\,y,\,z} \right)$ be points with integer coordinates satisfying the system of homogeneous equation:
$$\matrix{
{3x - y - z = 0} \cr
{ - 3x + z = 0} \cr
{ - 3x + 2y + z = 0} \cr
} $$
Then the number of such points for which $x^2 + {y^2} + {z^2} \le 100$ is
Correct Answer: 7
Explanation:
To solve this problem, we need to find the integer points $\left( {x,\,y,\,z} \right)$ that satisfy the given system of homogeneous equations:
$\matrix{ {3x - y - z = 0} \cr { - 3x + z = 0} \cr { - 3x + 2y + z = 0} \cr }$
Firstly, let’s solve for $z$ in terms of $x$ from the second equation:
$ - 3x + z = 0 \Rightarrow z = 3x $
Next, substitute $z = 3x$ into the first equation:
$ 3x - y - 3x = 0 \Rightarrow -y = 0 \Rightarrow y = 0 $
With $y = 0$ and $z = 3x$, the third equation also should be satisfied. Let's substitute $y$ and $z$ back into the third equation to verify:
This equation holds true, confirming that the solutions for $y$ and $z$ remain consistent. Therefore, the points that satisfy the given system are of the form:
$\left( x,\,0,\,3x \right)$
Additionally, we need $x^2 + y^2 + z^2 \le 100$. Substituting $y = 0$ and $z = 3x$, we get:
$ x^2 + 0^2 + (3x)^2 \le 100 $
This further simplifies to:
$ x^2 + 9x^2 \le 100 $
$ 10x^2 \le 100 $
$ x^2 \le 10 $
Hence, $ -\sqrt{10} \le x \le \sqrt{10} $
Since $x$ must be an integer, we evaluate acceptable values for $x$:
$x \in \{-3,\,-2,\,-1,\,0,\,1,\,2,\,3\}$
For each of these values, let’s determine the corresponding points $\left( x,\,0,\,3x \right)$:
$( -3,\,0,\,-9 )$
$( -2,\,0,\,-6 )$
$( -1,\,0,\,-3 )$
$( 0,\,0,\,0 )$
$( 1,\,0,\,3 )$
$( 2,\,0,\,6 )$
$( 3,\,0,\,9 )$
Thus, there are a total of 7 such points.
Therefore, the number of integer-coordinate points $\left( x,\,y,\,z \right)$ satisfying the given system of equations and the condition $x^2 + y^2 + z^2 \le 100$ is 7.
An n-digit number is a positive number with exactly digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is
