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JEE Advanced · Mathematics · Permutations and Combinations

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1
The number of ways to distribute 10 identical red pens and 14 identical blue pens among four persons such that each person gets exactly 6 pens, is
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Your answer: —
Correct answer: 206
2
Let SS be the set of all seven-digit numbers that can be formed using the digits 0,10, 1 and 22. For example, 22102222210222 is in SS, but 02102220210222 is NOT in SS. Then the number of elements xx in SS such that at least one of the digits 00 and 11 appears exactly twice in xx, is equal to ____________.
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Your answer: —
Correct answer: 762
3
A group of 99 students, s1,s2,…,s9s_1,s_2,\ldots,s_9, is to be divided to form three teams X,Y,ZX,Y,Z of sizes 2,3,2,3, and 44, respectively. Suppose that s1s_1 cannot be selected for the team XX, and s2s_2 cannot be selected for the team YY. Then the number of ways to form such teams, is _______.
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Your answer: —
Correct answer: 665
4
Let XX be the set of all five-digit numbers formed using 1,2,2,2,4,4,01,2,2,2,4,4,0 (each digit used at most as many times as it appears here). For example, 22240∈X22240\in X while 0224402244 and 4442244422 are not in XX. Suppose each element of XX has an equal chance of being chosen. Let pp be the conditional probability that an element chosen at random is a multiple of 2020 given that it is a multiple of 55. Then the value of 38p38p is equal to
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Your answer: —
Correct answer: 31
5
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?
Not attempted
A
2181621816
B
8553685536
C
1209612096
D
156816156816
6
The number of 44-digit integers in the closed interval [2022,4482][2022,4482] formed by using the digits 0,2,3,4,6,70,2,3,4,6,7 is __________.
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Your answer: —
Correct answer: 569
7
Let S1={(i,j,k):i,j,k∈{1,2,…,10}}S_1 = \{(i,j,k) : i,j,k \in \{1,2,\ldots,10\}\}, S2={(i,j):1≤i<j+2≤10, i,j∈{1,2,…,10}}S_2 = \{(i,j) : 1 \le i < j+2 \le 10,\ i,j \in \{1,2,\ldots,10\}\}, S3={(i,j,k,l):1≤i<j<k<l, i,j,k,l∈{1,2,…,10}}S_3 = \{(i,j,k,l) : 1 \le i < j < k < l,\ i,j,k,l \in \{1,2,\ldots,10\}\} and S4={(i,j,k,l):i,j,k,l are distinct elements of {1,2,…,10}}S_4 = \{(i,j,k,l) : i,j,k,l \text{ are distinct elements of } \{1,2,\ldots,10\}\}. If the total number of elements in the set SrS_r is nrn_r, r=1,2,3,4r=1,2,3,4, then which of the following statements is (are) TRUE?
Not attempted
A
n1=1000n_1 = 1000
B
n2=44n_2 = 44
C
n3=220n_3 = 220
D
n412=420\frac{n_4}{12} = 420
8
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 ____
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Your answer: —
Correct answer: 495
9
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 ____
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Your answer: —
Correct answer: 1080
10
Five persons A,B,C,DA,B,C,D and EE 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 ____
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Your answer: —
Correct answer: 30
11
In a high school, a committee has to be formed from a group of 66 boys M1,M2,M3,M4,M5,M6M_1,M_2,M_3,M_4,M_5,M_6 and 55 girls G1,G2,G3,G4,G5G_1,G_2,G_3,G_4,G_5. (i) Let α1\alpha_1 be the total number of ways in which the committee can be formed such that the committee has 55 members, having exactly 33 boys and 22 girls. (ii) Let α2\alpha_2 be the total number of ways in which the committee can be formed such that the committee has at least 22 members, and having an equal number of boys and girls. (iii) Let α3\alpha_3 be the total number of ways in which the committee can be formed such that the committee has 55 members, at least 22 of them being girls. (iv) Let α4\alpha_4 be the total number of ways in which the committee can be formed such that the committee has 44 members, having at least 22 girls and such that both M1M_1 and G1G_1 are NOT in the committee together. Match each entry in LIST-I to the correct entry in LIST-II. LIST-I\textbf{LIST-I} P. The value of α1\alpha_1 is Q. The value of α2\alpha_2 is R. The value of α3\alpha_3 is S. The value of α4\alpha_4 is LIST-II\textbf{LIST-II} 1. 136136 2. 189189 3. 192192 4. 200200 5. 381381 6. 461461
Not attempted
A
P→\to4; Q→\to6; R→\to2; S→\to1
B
P→\to1; Q→\to4; R→\to2; S→\to3
C
P→\to4; Q→\to6; R→\to5; S→\to2
D
P→\to4; Q→\to2; R→\to3; S→\to1
12
The number of 5 digit numbers which are divisible by 4, with digits from the set {1,2,3,4,5}\{1,2,3,4,5\} and the repetition of digits is allowed, is ______.
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Your answer: —
Correct answer: 625
13
How many 3×33\times 3 matrices MM with entries from {0,1,2}\{0, 1, 2\} are there, for which the sum of the diagonal entries of MTMM^T M is 55?
Not attempted
A
126126
B
198198
C
162162
D
135135
14
Let S={1,2,3,…,9}S = \{1, 2, 3, \ldots, 9\}. For k=1,2,…,5k = 1, 2, \ldots, 5, let NkN_k be the number of subsets of SS, each containing five elements, out of which exactly kk are odd. Then N1+N2+N3+N4+N5=N_1 + N_2 + N_3 + N_4 + N_5 =
Not attempted
A
210210
B
252252
C
125125
D
126126
15
Words of length 1010 are formed using the letters A,B,C,D,E,F,G,H,I,JA,B,C,D,E,F,G,H,I,J. Let xx be the number of such words where no letter is repeated, and let yy be the number of such words where exactly one letter is repeated twice and no other letter is repeated. Then y9x=\dfrac{y}{9x}=
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Your answer: —
Correct answer: 5
16
A debate club consists of 6 girls and 4 boys. A team of 4 members is to be selected 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
Not attempted
A
380
B
320
C
260
D
95
17
The coefficient of x9x^9 in the expansion of (1+x)(1+x2)(1+x3)…(1+x100)(1+x)(1+x^2)(1+x^3)\ldots(1+x^{100}) is
Not attempted
Your answer: —
Correct answer: 8
18
Let nn be the number of ways in which 55 boys and 55 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let mm be the number of ways in which 55 boys and 55 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue. Then the value of mn\dfrac{m}{n} is
Not attempted
Your answer: —
Correct answer: 5
19
Six cards and six envelopes are numbered 1,2,3,4,5,61,2,3,4,5,6, 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 11 is always placed in envelope numbered 22. Then the number of ways it can be done is
Not attempted
A
264264
B
265265
C
5353
D
6767
20
Let n≥2n\geq 2 be an integer. Take nn 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 blue and the rest red. If the number of red and blue line segments are equal, then the value of nn is
Not attempted
Your answer: —
Correct answer: 5
21
Let n1<n2<n3<n4<n5n_1<n_2<n_3<n_4<n_5 be positive integers such that n1+n2+n3+n4+n5=20n_1+n_2+n_3+n_4+n_5=20. Then the number of such distinct arrangements (n1,n2,n3,n4,n5)(n_1,n_2,n_3,n_4,n_5) is
Not attempted
Your answer: —
Correct answer: 7
Tip\textbf{Tip}: Substitute ni=ai+(i−1)n_i=a_i+(i-1) to convert the strictly increasing sequence into a non-decreasing one a1≤a2≤⋯≤a5a_1\leq a_2\leq\dots\leq a_5 of positive integers, turning the count into a partition-counting problem.
Deconstruction\textbf{Deconstruction}: Count strictly increasing 55-tuples of positive integers summing to 2020. Strategy\textbf{Strategy}: Let ni=ai+(i−1)n_i=a_i+(i-1) for i=1,…,5i=1,\dots,5, so strict inequality among nin_i corresponds exactly to a1≤a2≤⋯≤a5a_1\leq a_2\leq\dots\leq a_5, each ai≥1a_i\geq1. Execution\textbf{Execution}: ∑ni=∑ai+(0+1+2+3+4)=∑ai+10=20⇒∑ai=10\sum n_i = \sum a_i + (0+1+2+3+4) = \sum a_i+10=20 \Rightarrow \sum a_i=10, with 1≤a1≤⋯≤a51\leq a_1\leq \dots\leq a_5. This count equals the number of partitions of 1010 into exactly 55 positive parts, which equals the number of partitions of 55 into at most 55 parts, i.e. simply the number of partitions of 55: 5; 4+1; 3+2; 3+1+1; 2+2+1; 2+1+1+1; 1+1+1+1+15;\ 4{+}1;\ 3{+}2;\ 3{+}1{+}1;\ 2{+}2{+}1;\ 2{+}1{+}1{+}1;\ 1{+}1{+}1{+}1{+}1 — that is 77 partitions. Verification\textbf{Verification}: Translating back, a=(1,1,1,1,6)⇒n=(1,2,3,4,10)a=(1,1,1,1,6)\Rightarrow n=(1,2,3,4,10), sum =20=20; a=(2,2,2,2,2)⇒n=(2,3,4,5,6)a=(2,2,2,2,2)\Rightarrow n=(2,3,4,5,6), sum=20=20. Both check out. Final Answer\textbf{Final Answer}: 77.