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JEE • MATHEMATICS • PERMUTATION AND COMBINATION

Permutation and combination practice · Medium

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Practice format and chapter quiz

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 241-250 of 250 questions
241 Medium
241 of 250
Application +4 | -1
The number of permutations of 4 letters from the letters of word 'KOLKATA' is
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Correct option: B
270
Option B is correct because the number of permutations of 4 letters from the word 'KOLKATA' is 270, considering the repeated letters.
242 Medium
242 of 250
Application +4 | -1
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:
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Correct option: C
46th 46^{\text {th }}
Option C (46th) is correct because when arranging the letters of QUEEN in dictionary order, the word QUEEN appears after all words starting with E, N, QE, and QN.
243 Medium
243 of 250
Application +4 | -1
The value of 20Cn25Cm\frac{{ }^{20} C_n}{{ }^{25} C_m} when both numerator and denominator have their greatest values, is
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Correct option: C
1434025\frac{143}{4025}
Option C (1434025\frac{143}{4025}) is correct because it is the ratio of the greatest values of the binomial coefficients 20Cn{ }^{20} C_n and 25Cm{ }^{25} C_m, which occur at n=10n=10 and m=13m=13 respectively.
244 Medium
244 of 250
Analyzing +4 | -1
A class has 21 students. The class teacher has been asked to make nn groups of rr students each and go to zoo taking one group at a time. The size of group zoo i.e. the value of rr) for which the teacher goes to the maximum number of times is (no group can go to the zoo twice)
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Correct option: B
10 or 11
Option A (0,π4)∪(3π4,π)\left(0, \frac{\pi}{4}\right) \cup\left(\frac{3 \pi}{4}, \pi\right) is correct because it is the interval where β−α\beta - \alpha lies, derived from the given trigonometric conditions.
245 Medium
245 of 250
Application +4 | -1
The number of 3 digit numbers ' abca b c ' such that b<cb < c is
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Correct option: B
405
Option B (405) is correct because there are 9 choices for the first digit and then the number of ways to choose the second and third digits such that b < c is the sum of integers from 1 to 9.
246 Medium
246 of 250
Application +4 | -1
Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty?
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Correct option: C
150
Option C (150) is correct because it accounts for all possible distributions of the five balls into the three boxes such that no box is empty.
247 Medium
247 of 250
Application +4 | -1
The number of arrangements that can be formed from the letters a,b,c,d,e,fa, b, c, d, e, f taken 3 at a time without repetition and each arrangement containing at least one vowel, is
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Correct option: A
96
Option A (96) is correct because it correctly calculates the number of arrangements with at least one vowel by considering cases with one vowel and two consonants, and two vowels and one consonant.
248 Medium
248 of 250
Application +4 | -1
Let A={x1,x2,x3,  x4,  x5,  x6}A = \left\{ {{x_1},{x_{2,}}{x_3},\;{x_4},\;{x_5},\;{x_6}} \right\},
B={y1,  y2,  y3,y4,y5,y6}.B = \left\{ {{y_1},\;{y_2},\;{y_3},{y_4},{y_5},{y_6}} \right\}. . Then the number of one – one mapping from AA to BB such that f(xi)≠yvi=1,  2,  3,  4,  5,  6f\left( {{x_i}} \right) \ne {y_v}i = 1,\;2,\;3,\;4,\;5,\;6
View answer and explanation
Correct option: B
265
The correct answer is 265 because it is calculated by using the principle of inclusion-exclusion to find the number of one-to-one mappings where no element maps to itself.
249 Medium
249 of 250
Understanding +4 | -1
In a country, no two persons have identical set of teeth and there is no person without a tooth, also no person has more than 32 teeth. If shape and size of tooth is disregarded and only the position of tooth is considered, then maximum population of that country can be
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Correct option: B
232−12^{32}-1
Option B (232−12^{32}-1) is correct because each tooth can be either present or absent, leading to 2322^{32} possibilities, but we exclude the case where no tooth is present.
250 Medium
250 of 250
Application +4 | -1
The number of ways in which the letters of 'ARRANGE' be arranged so that i. "No two R's come together is
View answer and explanation
Correct option: B
900
Option B (900) is correct because it calculates the arrangements where no two R's are together by arranging the other letters first and then placing the R's in the gaps.
Revision notes and key takeaways

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