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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Easy

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 121-150 of 485 questions

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121 Easy
121 of 485
Application +4 | -1
The number of ways in which 6 men and 5 women can dine at a round table, if no two women are to sit together, is
View answer and explanation
Correct option: A
6!×5!6! \times 5!
Option A, 6!×5!6! \times 5!, is correct because we first arrange the men around the table and then arrange the women in the spaces between them.
122 Easy
122 of 485
Application +4 | -1
The sum of all 4 digited numbers that can be formed using the digits 1,2,4,5,61,2,4,5,6 without repetition is
View answer and explanation
Correct option: A
479952
Option A (479952) is correct because it calculates the sum of all 4-digit numbers formed using the given digits without repetition.
123 Easy
123 of 485
Application +4 | -1
If   12Pr=  11P6+6.  11P5{\;^{12}}{P_r} = {\;^{11}}{P_6} + 6.{\;^{11}}{P_5} then rr is equal to
View answer and explanation
Correct option: A
6
Option A (6) is correct because by using the given relation and simplifying, we find that r = 6.
124 Easy
124 of 485
Application +4 | -1
The number of lines drawn through 6 points lying on a circle is
View answer and explanation
Correct option: B
15
Option B (15) is correct because it calculates the number of ways to choose 2 points out of 6 to form a line.
125 Easy
125 of 485
Understanding +4 | -1
How many 10 digits numbers can be written by using digits 9 and 2?
View answer and explanation
Correct option: B
210{2^{10}}
Option B (210{2^{10}}) is correct because each of the 10 digits can be either a 9 or a 2, giving 2 options for each digit.
126 Easy
126 of 485
Application +4 | -1
Total number of three letter words that can be formed from the letters of the word 'SAHARANPUR' is equal to
View answer and explanation
Correct option: C
247
Option C (247) is correct because it accurately counts the number of three-letter words that can be formed from the letters of 'SAHARANPUR'.
127 Easy
127 of 485
Application +4 | -1
In a circus there are ten cages for accommodating ten animals. Out of these four cages are so small that five out of 10 animals cannot enter into them. In how many ways will it be possible to accommodate ten animals in these ten cages?
View answer and explanation
Correct option: B
86400
Option B (86400) is correct because it first arranges the animals that can't go into the small cages, then arranges the remaining animals in the remaining cages.
128 Easy
128 of 485
Application +4 | -1
If the permutations of a,b,c,d,ea,b,c,d,e taken all together be written down in alphabetical order as in dictionary and numbered, then the rank of the permutation debacdebac is
View answer and explanation
Correct option: D
93
Option D (93) is correct because the rank of the permutation 'debac' is determined by counting the permutations that come before it in lexicographical order.
129 Easy
129 of 485
Analyzing +4 | -1
Let pp be a prime number such that p≥3p \geq 3. Let n=p!+1.n=p !+1. The number of primes in the list n+1,n+2,n+3,....,n+p−1n + 1, n + 2, n + 3, ...., n + p − 1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because for each kk in the list, n+kn+k is divisible by k+1k+1, meaning none of the numbers in the list are prime.
130 Easy
130 of 485
Application +4 | -1
How many numbers divisible by 5 and lying between 3000 and 4000 that can be formed from the digits 1, 2, 3, 4, 5, 6 (repetition of digits is not allowed)?
View answer and explanation
Correct option: C
  4P2{\;^4}{P_2}
Option A (216) is incorrect. Option B (375) is correct because it correctly calculates the total number of four-digit odd numbers that can be formed using the given digits.
131 Easy
131 of 485
Application +4 | -1
If [y][y] denote the greatest integer ≤y\leq y, and 2[x8]2+3[x8]=202\left[\frac{x}{8}\right]^2+3\left[\frac{x}{8}\right]=20, then xx lies in the smallest interval [a,b)[a, b) where b−ab-a is equal to
View answer and explanation
Correct option: D
8
Option D (LICPEN) is correct because by arranging the permutations in dictionary order, the 413th word is LICPEN.
132 Easy
132 of 485
Application +4 | -1
How many different arrangements can be made out of the letters in the expansion A2B3C4A^2 B^3 C^4, when written in full?
View answer and explanation
Correct option: B
9!2!3!4!\frac{9 !}{2 ! 3 ! 4 !}
Option B (9!2!3!4!\frac{{9 !}}{{2 ! 3 ! 4 !}}) is correct because it correctly applies the formula for permutations with repetitions.
133 Easy
133 of 485
Application +4 | -1
If   nPr=30240{\;^n}{P_r} = 30240 and   nCr=252{\;^n}{C_r} = 252, then the ordered pair (n,  r)\left( {n,\;r} \right) is equal to
View answer and explanation
Correct option: B
(10,5)
Option B (10,5) is correct because it satisfies both the permutation and combination equations given in the problem.
134 Easy
134 of 485
Application +4 | -1
In an examination a minimum is to be secured in each of 5 subjects for a pass. The number of ways in which a student can fail is
View answer and explanation
Correct option: A
31
Option A (31) is correct because it represents the total number of ways a student can fail in at least one subject out of 5, calculated using combinations.
135 Easy
135 of 485
Application +4 | -1
A1,A2…AnA_1, A_2 \ldots A_n are points on the line y=xy=x lying in the positive quadrant such that OAn=nOAn−1,OO A_n=n O A_{n-1}, O being the origin. If OA1=1O A_1=1 and the coordinates of AnA_n are (25202,25202)(2520 \sqrt{2}, 2520 \sqrt{2}), then n=n=
View answer and explanation
Correct option: C
7
Option C (7) is correct because OAn=n!OA_n = n! and solving for nn with the given coordinates of AnA_n gives n=7n = 7.
136 Easy
136 of 485
Application +4 | -1
Let Pm{P_m} stand for   mPm.{\;^m}{P_m}. Then, 1+P1+2  P2+3  P3+…+n.Pn1 + {P_1} + 2\;{P_2} + 3\;{P_3} + \ldots + n.{P_n} is equal to
View answer and explanation
Correct option: C
(n+1)  !−1\left( {n + 1} \right)\;! - 1
Option C ((n+1) !−1\left( {n + 1} \right)\,! - 1) is correct because it represents the sum of the series 1+P1+2  P2+3  P3+…+n.Pn1 + {P_1} + 2\;{P_2} + 3\;{P_3} + \ldots + n.{P_n}.
137 Easy
137 of 485
Application +4 | -1
The number of ways in which the letters of the word "STRANGE" can be arranged so that the vowels may appear in the odd places, is
View answer and explanation
Correct option: A
1440
Option A (1440) is correct because the 2 vowels can be arranged in 4 odd places in P(4,2){P}(4,2) ways and the 5 consonants can be arranged in the remaining 5 places in P(5,5){P}(5,5) ways.
138 Easy
138 of 485
Application +4 | -1
The number of all possible words that can be formed using the letters of the word “MATHEMATICS” is
View answer and explanation
Correct option: A
11  !2  !2  !2  !\frac{{11\;!}}{{2\;!2\;!2\;!}}
Option A (11  !2  !2  !2  !\frac{{11\;!}}{{2\;!2\;!2\;!}}) is correct because it accounts for the repetitions of the letters M, A, and T in the word MATHEMATICS.
139 Easy
139 of 485
Application +4 | -1
Find the number of ways of arranging the letters of the word 'BANANA'
View answer and explanation
Correct option: A
60
Option A (60) is correct because it accounts for the repetitions of the letters A and N in the word BANANA.
140 Easy
140 of 485
Application +4 | -1
The number of ways in which 20 different pearls of two colours can be set alternately on a necklace, there being 10 pearls of each colour, is
View answer and explanation
Correct option: B
5×(9!)25 \times {\left( {9!} \right)^2}
Option B (5×(9!)25 \times {\left( {9!} \right)^2}) is correct because it accounts for arranging the pearls of one color in a circle and then arranging the pearls of the other color in the spaces.
141 Easy
141 of 485
Application +4 | -1
Let AA be a set of n(≥3)n(\geq 3) distinct elements. The number of triplets (x,y,z)(x, y, z) of the set AA in which atleast two coordinates are equal to
View answer and explanation
Correct option: B
n3−nP3n^3-{ }^n P_3
Option B (n3−nP3n^3-{ }^n P_3) is correct because it calculates the total number of triplets minus the number of triplets with all distinct elements.
142 Easy
142 of 485
Application +4 | -1
Out of 18 points in a plane no three are in the same straight line except five points which are collinear. The number of straight lines that can be formed joining them is
View answer and explanation
Correct option: B
144
Option B (144) is correct because it calculates the total number of lines minus the extra lines counted due to collinearity, plus one for the single line formed by the collinear points.
143 Easy
143 of 485
Application +4 | -1
Let TnT_n be the number of all possible triangles formed by joining vertices of an nn-sided regular polygon. If Tn+1−Tn=10T_{n+1}-T_n=10, the value of nn is
View answer and explanation
Correct option: A
5
Option A (5) is correct because the equation Tn+1−Tn=10{T_{n + 1}} - {T_n} = 10 simplifies to nC2=10{^n}{C_2} = 10, which is satisfied when n = 5.
144 Easy
144 of 485
Application +4 | -1
A seven-digit number made up of all distinct digits 8,7,6,4,2,x8,7,6,4,2, x and yy is divisible by 3. Then, possible number of order pair (x,y)(x, y) is
View answer and explanation
Correct option: B
8
Option B (8) is correct because there are 8 possible ordered pairs (x, y) that make the seven-digit number divisible by 3.
145 Easy
145 of 485
Application +4 | -1
If r,  s,  tr,\;s,\;t are prime numbers and p,qp,q are the positive integers such that LCM of p,  qp,\;q is r2s4t2{r^2}{s^4}{t^2}, then the number of ordered pairs (p,  q)\left( {p,\;q} \right) is
View answer and explanation
Correct option: C
225
Option C (225) is correct because the number of ordered pairs (p, q) is found by considering all possible combinations of the prime factors' exponents in the LCM.
146 Easy
146 of 485
Application +4 | -1
How many numbers greater than 40000 can be formed from the digits 2, 4, 5, 5, 7?
View answer and explanation
Correct option: D
48
Option D (48) is correct because it is calculated by subtracting the number of arrangements with 2 at the first place from the total number of arrangements.
147 Easy
147 of 485
Application +4 | -1
How many words can be made from the letters of the word ‘COMMITTEE’?
View answer and explanation
Correct option: B
9!(2!)3!\frac{{9!}}{{{{\left( {2!} \right)}^3}!}}
Option B (9!(2!)3!\frac{{9!}}{{{{\left( {2!} \right)}^3}!}}) is correct because it accounts for the repeated letters (M, T, E) in the word COMMITTEE.
148 Easy
148 of 485
Application +4 | -1
Every one of the 10 available lamps can be switched on to illuminate certain Hall. The total number of ways in which the hall can be illuminated, is
View answer and explanation
Correct option: B
10231023
Option B, 10231023, is correct because each lamp can be either on or off, giving 2102^{10} possibilities, but we must subtract 1 for the case where all lamps are off.
149 Easy
149 of 485
Application +4 | -1
A four digit number of distinct digits is formed by using the digits 2,3,4,5,6,7,82,3,4,5,6,7,8. The number of such numbers which are divisible by 25, is
View answer and explanation
Correct option: B
40
Option B (40) is correct because it considers the two possible endings (25 and 75) and calculates the permutations for the remaining digits.
150 Easy
150 of 485
Application +4 | -1
The least positive integer nn for which n−1C5+n−1C6<nC7{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7 is
View answer and explanation
Correct option: A
14
Option A (14) is correct because it is the smallest positive integer that satisfies the inequality n−1C5+n−1C6<nC7{ }^{n-1} C_5+{ }^{n-1} C_6 < { }^n C_7, which simplifies to n > 13.
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