Permutations and Combinations
414 Questions
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 4th September Evening Slot
A test consists of 6 multiple choice questions, each having 4 alternative answers of which only one is correct. The number of ways, in which a candidate answers all six questions such that exactly four of the answers are correct, is __________.
Correct Answer: 135
Explanation:
Select any 4 questions in 6C4
ways which are
correct.
Answering right option for each question is possible in 1 way.
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4
Number of ways of choosing wrong option for each question = 3
So ways of choosing wrong option for 2 questions = (3)2
$ \therefore $ Required number of ways = 6C4.(1)4.(3)2 = 135
Answering right option for each question is possible in 1 way.
So ways of choosing right option for 4 questions = 1.1.1.1 = (1)4
Number of ways of choosing wrong option for each question = 3
So ways of choosing wrong option for 2 questions = (3)2
$ \therefore $ Required number of ways = 6C4.(1)4.(3)2 = 135
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 3rd September Evening Slot
The total number of 3-digit numbers, whose
sum of digits is 10, is __________.
Correct Answer: 54
Explanation:
Let xyz is 3 digits number.
Given that sum of digits = 10
$ \therefore $ x + y + z = 10 ......(1)
Also x can't be 0 as if x = 0 then it will become 2 digits number.
So, x $ \ge $ 1, y $ \ge $ 0, z $ \ge $ 0
As x $ \ge $ 1
$ \Rightarrow $ x $-$ 1 $ \ge $ 0
Let x $-$ 1 = t
$ \therefore $ t $ \ge $ 0
From equation (1)
(x $-$ 1) + y + z = 9
$ \Rightarrow $ t + y + z = 9
Now this problem becomes, distributing 9 things among 3 people t, y, z.
Number of ways we can do that
= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$
Now when 3 digit number is 900 then t = 9, y = 0, z = 0.
And when t = 9, then
x $-$ 1 = 9
$ \Rightarrow $ x = 10
But we can't take x = 10 in a 3 digits number. So, we have to remove this case.
$ \therefore $ Total number of 3 digit numbers = 55 $-$ 1 = 54.
Given that sum of digits = 10
$ \therefore $ x + y + z = 10 ......(1)
Also x can't be 0 as if x = 0 then it will become 2 digits number.
So, x $ \ge $ 1, y $ \ge $ 0, z $ \ge $ 0
As x $ \ge $ 1
$ \Rightarrow $ x $-$ 1 $ \ge $ 0
Let x $-$ 1 = t
$ \therefore $ t $ \ge $ 0
From equation (1)
(x $-$ 1) + y + z = 9
$ \Rightarrow $ t + y + z = 9
Now this problem becomes, distributing 9 things among 3 people t, y, z.
Number of ways we can do that
= ${}^{9 + 3 - 1}{C_{3 - 1}} = {}^{11}{C_2} = 55$
Now when 3 digit number is 900 then t = 9, y = 0, z = 0.
And when t = 9, then
x $-$ 1 = 9
$ \Rightarrow $ x = 10
But we can't take x = 10 in a 3 digits number. So, we have to remove this case.
$ \therefore $ Total number of 3 digit numbers = 55 $-$ 1 = 54.
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 2nd September Morning Slot
If the letters of the word 'MOTHER' be permuted
and all the words so formed (with or without
meaning) be listed as in a dictionary, then the
position of the word 'MOTHER' is ______.
Correct Answer: 309
Explanation:
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Evening Slot
The number of 4 letter words (with or without
meaning) that can be formed from the eleven
letters of the word 'EXAMINATION' is
_______.
Correct Answer: 2454
Explanation:
2A, 2I, 2N, E, X, M, T, O
To form four letter words
Case 1 : All same ( not possible)
Case 2 : 1 different, 3 same (not possible)
Case 3 : 2 different, 2 same
= 3C1 $ \times $ 7C2 $ \times $ ${{4!} \over {2!}}$ = 756
Case 4 : 2 same of one kind, 2 same same of other kind
= 3C2 $ \times $ ${{4!} \over {2!2!}}$ = 18
Case 5 : All letters are different
= 8C4 $ \times $ 4! = 1680
$ \therefore $ Total ways = 1680 + 756 + 18 = 2454
To form four letter words
Case 1 : All same ( not possible)
Case 2 : 1 different, 3 same (not possible)
Case 3 : 2 different, 2 same
= 3C1 $ \times $ 7C2 $ \times $ ${{4!} \over {2!}}$ = 756
Case 4 : 2 same of one kind, 2 same same of other kind
= 3C2 $ \times $ ${{4!} \over {2!2!}}$ = 18
Case 5 : All letters are different
= 8C4 $ \times $ 4! = 1680
$ \therefore $ Total ways = 1680 + 756 + 18 = 2454
2020
JEE Mains
Numerical
JEE Main 2020 (Online) 8th January Morning Slot
An urn contains 5 red marbles, 4 black marbles
and 3 white marbles. Then the number of ways
in which 4 marbles can be drawn so that at the
most three of them are red is ___________.
Correct Answer: 490
Explanation:
Here 5 red marbels and 7 non red marbels presents.
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
Case 1: When 3 red marbels present
No of ways = 5C3 $ \times $ 7C1
Case 2: When 2 red marbels present
No of ways = 5C2 $ \times $ 7C2
Case 3: When 1 red marbels present
No of ways = 5C1 $ \times $ 7C3
Case 4: When 0 red marbels present
No of ways = 5C0 $ \times $ 7C4
$ \therefore $ Total number of ways
= 5C3 $ \times $ 7C1 + 5C2 $ \times $ 7C2 + 5C1 $ \times $ 7C3 + 5C0 $ \times $ 7C4
= 70 + 210 + 175 + 35
= 490
No of ways 4 marbels can be chosen where atmost 3 red marbels can be present.
Case 1: When 3 red marbels present
No of ways = 5C3 $ \times $ 7C1
Case 2: When 2 red marbels present
No of ways = 5C2 $ \times $ 7C2
Case 3: When 1 red marbels present
No of ways = 5C1 $ \times $ 7C3
Case 4: When 0 red marbels present
No of ways = 5C0 $ \times $ 7C4
$ \therefore $ Total number of ways
= 5C3 $ \times $ 7C1 + 5C2 $ \times $ 7C2 + 5C1 $ \times $ 7C3 + 5C0 $ \times $ 7C4
= 70 + 210 + 175 + 35
= 490
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 6th September Morning Slot
Two families with three members each and one family with four members are to be seated in a row.
In how many ways can they be seated so that the same family members are not separated?
A.
2! 3! 4!
B.
(3!)3.(4!)
C.
3! (4!)3
D.
(3!)2.(4!)
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 5th September Evening Slot
There are 3 sections in a question paper and
each section contains 5 questions. A candidate
has to answer a total of 5 questions, choosing
at least one question from each section. Then
the number of ways, in which the candidate
can choose the questions, is :
A.
2250
B.
2255
C.
3000
D.
1500
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 3rd September Morning Slot
The value of (2.1P0
– 3.2P1 + 4.3P2 .... up to
51th term)
+ (1! – 2! + 3! – ..... up to 51th term) is equal to :
+ (1! – 2! + 3! – ..... up to 51th term) is equal to :
A.
1
B.
1 + (51)!
C.
1 – 51(51)!
D.
1 + (52)!
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 2nd September Evening Slot
Let n > 2 be an integer. Suppose that there are
n Metro stations in a city located along a
circular path. Each pair of stations is connected
by a straight track only. Further, each pair of
nearest stations is connected by blue line,
whereas all remaining pairs of stations are
connected by red line. If the number of red lines
is 99 times the number of blue lines, then the
value of n is :
A.
201
B.
199
C.
101
D.
200
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 9th January Morning Slot
If the number of five digit numbers with distinct
digits and 2 at the 10th place is 336 k, then k
is equal to :
A.
6
B.
8
C.
4
D.
7
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 8th January Morning Slot
If a, b and c are the greatest value of 19Cp, 20Cq
and 21Cr respectively, then :
A.
${a \over {11}} = {b \over {22}} = {c \over {21}}$
B.
${a \over {10}} = {b \over {22}} = {c \over {21}}$
C.
${a \over {10}} = {b \over {11}} = {c \over {42}}$
D.
${a \over {11}} = {b \over {22}} = {c \over {42}}$
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Evening Slot
The number of ordered pairs (r, k) for which
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
6.35Cr = (k2 - 3). 36Cr + 1, where k is an integer, is :
A.
6
B.
3
C.
2
D.
4
2020
JEE Mains
MCQ
JEE Main 2020 (Online) 7th January Morning Slot
Total number of 6-digit numbers in which only and all the five digits 1, 3, 5, 7 and 9 appear, is :
A.
${5 \over 2}\left( {6!} \right)$
B.
${6!}$
C.
56
D.
${1 \over 2}\left( {6!} \right)$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
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
x1 + a + b + c + x5 = 8, and it is equal to
${}^{8 + 5 - 1}{C_{5 - 1}} = {}^{12}{C_4} = {{12 \times 11 \times 10 \times 9} \over {4 \times 3 \times 2}} = 495$
$ \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
x1 + a + b + c + x5 = 8, and it is equal to
${}^{8 + 5 - 1}{C_{5 - 1}} = {}^{12}{C_4} = {{12 \times 11 \times 10 \times 9} \over {4 \times 3 \times 2}} = 495$
2020
JEE Advanced
Numerical
JEE Advanced 2020 Paper 2 Offline
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
= $\eqalign{ & {{6!} \over {{{(2!)}^2}{{(1!)}^2}(2!)(2!)}}(4!) \cr & = 360 \times 3 = 1080 \cr} $
$ \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
= $\eqalign{ & {{6!} \over {{{(2!)}^2}{{(1!)}^2}(2!)(2!)}}(4!) \cr & = 360 \times 3 = 1080 \cr} $
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
For $n=1,2,3, \ldots .50$, let
$ A=\left\{a_n / a_n=\left\{\begin{array}{ll} (-1)^{\frac{n}{2}}\left(\frac{n}{2}\right), & \text { if } n \text { is even } \\ (-1)^{\frac{n-1}{2}}\left(\frac{n-1}{2}\right), & \text { if } n \text { is odd } \end{array}\right\}\right\} $
and $B$ is the set of all distinct elements of $A$. The number of permutations all the elements of set $B$ such that even integers are in increasing order, is
A.
$\frac{26!}{12!}$
B.
$\frac{49!}{12!13!}$
C.
$\frac{50!}{24!26!}$
D.
$\frac{26!}{13!12!}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
If $\alpha$ represents the number of arrangements of $p$ men and $q$ women in a row such that all men are together and $\beta$ represents the number of circular arrangements of the same people with the same condition, then $\alpha: \beta$ is
A.
$(q+1) p!: 1$
B.
$(q+1): 1$
C.
$1: p$ !
D.
$p!: q!$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
Consider the following statements:
I. The number of positive integral solutions of $x_1+x_2+x_3+x_4=10$ is 286 .
II. If $25!=10^n \times k,(k \in \mathbf{N})$, then $n=6$
Which one of the following options is true?
A.
Only I is true
B.
Only II is true
C.
Both I and II are true
D.
Both I and II are false
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 14th September Evening Shift
A student is allowed to select at least $(n+1)$ books but not all books from a collection of ( $2 n+1$ ) books. If the total number of ways in which he can select these books is 255 , then the number of books in that collection is
A.
4
B.
9
C.
10
D.
7
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
If $x$ and $y$ represent the number of arrangements of the letters of word ATRAPATRAM such that (i) all A's are together and (ii) no two A's are together respectively, then $x+y$
A.
$\frac{10!}{4!2!2!}$
B.
$\frac{7!\times 15}{2!2!4!}$
C.
$\frac{6!}{2!2!} \times 42$
D.
$\frac{7!}{2!2!}+\frac{6!\cdot 7 p_4}{2!2!}$
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Evening Shift
Numbers between 1 and 10,000 are formed using the digits 2 and 3 only once and the digit 4 twice. If the numbers thus formed are arranged in increasing order and $x, y$ represent the ranks of 4324 and 324 respectively then $x-y=$
A.
17
B.
31
C.
14
D.
16
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
The total number of three digit and five digit integers which can be formed by using the digits $0,1,2,3,4,5$ but using each digit not more than once in each number is
A.
100
B.
600
C.
700
D.
800
2020
TS-EAMCET
MCQ
TS EAMCET 2020 (Online) 10th September Morning Shift
At an election a voter may vote for any number of candidates not exceeding the number to be elected. If 4 candidates are to be elected out of the 12 contested in the election and voter votes for at least one candidate, then the number of ways in which a voter can vote is
A.
793
B.
298
C.
781
D.
1585
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Evening Slot
A group of students comprises of 5 boys and n girls. If the number of ways, in which a team of 3 students can
randomly be selected from this group such that there is at least one boy and at least one girl in each team, is
1750, then n is equal to :
A.
24
B.
25
C.
27
D.
28
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th April Morning Slot
The number of ways of choosing 10 objects out of 31 objects of which 10 are identical and the remaining 21
are distinct, is :
A.
220 - 1
B.
220
C.
220 + 1
D.
221
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Evening Slot
Suppose that 20 pillars of the same height have been erected along the boundary of a circular stadium. If the
top of each pillar has been connected by beams with the top of all its non-adjacent pillars, then the total
number of beams is :
A.
180
B.
210
C.
170
D.
190
Any two non-adjacent pillers are joined by beams
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th April Morning Slot
The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which are divisible by
11 and no digit is repeated is :
A.
36
B.
60
C.
72
D.
48
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th April Morning Slot
A committee of 11 members is to be formed from
8 males and 5 females. If m is the number of ways
the committee is formed with at least 6 males and
n is the number of ways the committee is formed
with at least 3 females, then :
A.
n = m – 8
B.
m = n = 78
C.
m + n = 68
D.
m = n = 68
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Evening Slot
The number of four-digit numbers strictly greater
than 4321 that can be formed using the digits
0,1,2,3,4,5 (repetition of digits is allowed) is :
A.
306
B.
288
C.
310
D.
360
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 8th April Morning Slot
All possible numbers are formed using the digits
1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number
of such numbers in which the odd digits occupy
even places is :
A.
175
B.
162
C.
160
D.
180
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Evening Slot
There are m men and two women participating in a chess tournament. Each participant plays two games with every other participant. If the number of games played by the men between themselves exceeds the number of games played between the men and the women by 84, then the value of m is :
A.
12
B.
9
C.
7
D.
11
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 12th January Morning Slot
Consider three boxes, each containing, 10 balls labelled 1, 2, … , 10. Suppose one ball is randomly drawn from each of the boxes. Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3). Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is :
A.
164
B.
240
C.
82
D.
120
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 10th January Evening Slot
If $\sum\limits_{r = 0}^{25} {\left\{ {{}^{50}{C_r}.{}^{50 - r}{C_{25 - r}}} \right\} = K\left( {^{50}{C_{25}}} \right)} ,\,\,$ then K is equal to :
A.
224
B.
225$-$ 1
C.
225
D.
(25)2
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
Let S be the set of all triangles in the xy-plane, each having one vertex at the origin and the other two vertices lie on coordinate axes with integral coordinates. If each triangle in S has area 50 sq. units, then the number of elements in the set S is :
A.
9
B.
18
C.
36
D.
32
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Evening Slot
The number of natural numbers less than 7,000 which can be formed by using the digits 0, 1, 3, 7, 9 (repitition of digits allowed) is equal to :
A.
374
B.
372
C.
375
D.
250
2019
JEE Mains
MCQ
JEE Main 2019 (Online) 9th January Morning Slot
Consider a class of 5 girls and 7 boys. The number of different teams consisting of 2 girls and 3 boys that can
be formed from this class, if there are two specific boys A and B, who refuse to be the members of the same
team, is :
A.
500
B.
350
C.
200
D.
300
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 2 Offline
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
= 6C5[ 6C6]
So, total number of ways of choosing sets A and B
= 6C1[6C2 + 6C3 + 6C4 + 6C5 + 6C6]
+ 6C2[ 6C3 + 6C4 + 6C5 + 6C6]
+ 6C3[ 6C4 + 6C5 + 6C6]
+ 6C4[ 6C5 + 6C6]
+ 6C5[ 6C6]
= Sum of all possible products of two terms from
6C1, 6C2, 6C3, ......., 6C6 = p(Assume)
Now we know,
(6C1 + 6C2 + .....+ 6C6)2
= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2[6C16C2 + 6C16C3 + ... +6C16C6
+ 6C26C3 + 6C26C4 + ... +6C26C6+ ....+ 6C56C6]
$ \Rightarrow $ (6C1 + 6C2 + .....+ 6C6)2
= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2(p)
$ \Rightarrow $ (26 - 6C0)2 = [ 12C6 - (6C0)2 ] + 2p
$ \Rightarrow $ p $ = {{{{({2^6} - 1)}^2} - ({}^{12}{C_6} - 1)} \over 2}$
$ = {{{{(63)}^2} - {{12!} \over {6!6!}} + 1} \over 2}$
$ = {{3969 - 924 + 1} \over 2} = {{3046} \over 2} = 1523$
Note :
(nC0 + nC1 + .....+ nCn)2
= (nC0)2 + (nC1)2+ ....+ (nCn)2 + 2[ nC0nC1 + nC0nC2 + ... +nC0nCn
+ nC1nC2 + nC1nC3 + ... + nC1nCn
+ nC2nC3 + nC2nC4 + ... +nC2nCn+ ....+ nCn - 1nCn]
$ \Rightarrow $ (2n)2 = 2nCn + 2(Sum of all possible products of two terms from
nC1, nC2, nC3, ......., nCn)
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
= 6C5[ 6C6]
So, total number of ways of choosing sets A and B
= 6C1[6C2 + 6C3 + 6C4 + 6C5 + 6C6]
+ 6C2[ 6C3 + 6C4 + 6C5 + 6C6]
+ 6C3[ 6C4 + 6C5 + 6C6]
+ 6C4[ 6C5 + 6C6]
+ 6C5[ 6C6]
= Sum of all possible products of two terms from
6C1, 6C2, 6C3, ......., 6C6 = p(Assume)
Now we know,
(6C1 + 6C2 + .....+ 6C6)2
= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2[6C16C2 + 6C16C3 + ... +6C16C6
+ 6C26C3 + 6C26C4 + ... +6C26C6+ ....+ 6C56C6]
$ \Rightarrow $ (6C1 + 6C2 + .....+ 6C6)2
= [ (6C1)2 + (6C2)2+ ....+ (6C6)2] + 2(p)
$ \Rightarrow $ (26 - 6C0)2 = [ 12C6 - (6C0)2 ] + 2p
$ \Rightarrow $ p $ = {{{{({2^6} - 1)}^2} - ({}^{12}{C_6} - 1)} \over 2}$
$ = {{{{(63)}^2} - {{12!} \over {6!6!}} + 1} \over 2}$
$ = {{3969 - 924 + 1} \over 2} = {{3046} \over 2} = 1523$
Note :
(nC0 + nC1 + .....+ nCn)2
= (nC0)2 + (nC1)2+ ....+ (nCn)2 + 2[ nC0nC1 + nC0nC2 + ... +nC0nCn
+ nC1nC2 + nC1nC3 + ... + nC1nCn
+ nC2nC3 + nC2nC4 + ... +nC2nCn+ ....+ nCn - 1nCn]
$ \Rightarrow $ (2n)2 = 2nCn + 2(Sum of all possible products of two terms from
nC1, nC2, nC3, ......., nCn)
2019
JEE Advanced
Numerical
JEE Advanced 2019 Paper 2 Offline
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 are2R + 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
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 16th April Morning Slot
The number of numbers between 2,000 and 5,000 that can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is not allowed) and are multiple of 3 is :
A.
24
B.
30
C.
36
D.
48
2018
JEE Mains
MCQ
JEE Main 2018 (Offline)
From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and
arranged in a row on a shelf so that the dictionary is always in the middle. The number of such
arrangements is :
A.
at least 750 but less than 1000
B.
at least 1000
C.
less than 500
D.
at least 500 but less than 750
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Evening Slot
The number of four letter words that can be formed using the letters of the word BARRACK is :
A.
120
B.
144
C.
264
D.
270
2018
JEE Mains
MCQ
JEE Main 2018 (Online) 15th April Morning Slot
n$-$digit numbers are formed using only three digits 2, 5 and 7. The smallest value of n for which 900 such distinct numbers can be formed, is :
A.
6
B.
7
C.
8
D.
9
2018
JEE Advanced
MCQ
JEE Advanced 2018 Paper 2 Offline
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.
The correct option is
(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 |
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
2018
JEE Advanced
Numerical
JEE Advanced 2018 Paper 1 Offline
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
$ \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
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 9th April Morning Slot
The number of ways in which 5 boys and 3 girls can be seated on a round table if a
particular boy B1 and a particular girl G1 never sit adjacent to each other, is :
A.
5 $ \times $ 6!
B.
6 $ \times $ 6!
C.
7!
D.
5 $ \times $ 7!
2017
JEE Mains
MCQ
JEE Main 2017 (Online) 8th April Morning Slot
If all the words, with or without meaning, are written using the letters of the word QUEEN and are arranged as in English dictionary, then the position of the word QUEEN is :
A.
44th
B.
45th
C.
46th
D.
47th
2017
JEE Mains
MCQ
JEE Main 2017 (Offline)
A man X has 7 friends, 4 of them are ladies and 3 are men. His wife Y also has 7 friends, 3 of them are
ladies and 4 are men. Assume X and Y have no common friends. Then the total number of ways in which X
and Y together can throw a party inviting 3 ladies and 3 men, so that 3 friends of each of X and Y are in
this party, is:
A.
468
B.
469
C.
484
D.
485
2017
JEE Advanced
Numerical
JEE Advanced 2017 Paper 1 Offline
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!
Thus, ${y \over {9x}} = {{{}^{10}{C_1} \times {}^{10}{C_2} \times {}^9{C_8} \times 8!} \over {9 \times 10!}}$
Using ${}^n{C_r} = {{n!} \over {r!(n - r)!}}$, we get
${y \over {9x}} = {{\left[ {{{10!} \over {1!(10 - 1)!}} \times {{10!} \over {2!(10 - 2)!}} \times {{9!} \over {8!(9 - 8)!}} \times 8!} \right]} \over {9 \times 10!}}$
$ = {{10!} \over {9 \times 2!\, \times 8!}} = {{10 \times 9 \times 8!} \over {9 \times 2 \times 8!}} = {{10} \over 2} = 5$ [Using $n! = n(n - 1)(n - 2)......1!$]
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
If ${{{}^{n + 2}C{}_6} \over {{}^{n - 2}{P_2}}}$ = 11, then n satisfies the
equation :
A.
n2 + 3n − 108 = 0
B.
n2 + 5n − 84 = 0
C.
n2 + 2n − 80 = 0
D.
n2 + n − 110 = 0
2016
JEE Mains
MCQ
JEE Main 2016 (Online) 10th April Morning Slot
The sum $\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right) \times \left( {r!} \right)} $ is equal to :
A.
(11)!
B.
10 $ \times $ (11!)
C.
101 $ \times $ (10!)
D.
11 $ \times $ (11!)