InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Quantity 2: If nC5 = nC6 , then 13Cn = ?1). Quantity 2 > Quantity 12). Quantity 2 < Quantity 13). Quantity 2 ≥ Quantity 14). Quantity 2 ≤ Quantity 1 |
|
Answer» First we will find QUANTITY 1, Quantity 1: 10 × nC2 = 3 × n + 1C3 (? nCr = $(\frac{{n!}}{{\left( {n - r} \right)!\left( {r!} \right)}})$) $(\Rightarrow \;10\; \TIMES \;\frac{{n\left( {n - 1} \right)}}{{1\; \times \;2}}\; = \;3\; \times \;\frac{{\left( {n\; + \;1} \right)\;n\;\left( {n - 1} \right)}}{{1\; \times \;2\; \times \;3}})$ ⇒ 10 = n + 1 ⇒ n = 9 ∴ 9N = 9 × 9 = 81 Now, Quantity 2: nC5 = nC6 (? nCr = nCs then r = s or r + s = n) ⇒ n = 5 + 6 = 11 ∴ n = 11 13Cn = 13C11 = 13C2 $(= \;\frac{{13\; \times \;12}}{{1\; \times \;2}}\; = \;78)$ ∴ Quantity 1 > Quantity 2 |
|
| 2. |
The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m (m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is :1). 82). 123). 94). 10 |
|
Answer» If n is even, then the number of BOYS should be equal to number of girls, LET each be a So, n = 2a Then the number of arrangements = 2 × a! × a! If one more STUDENTS is added, then number of arrangements = a! × (a + 1)! But this is 200% more than the earlier Hence, 3(2 × a! × a!) = a! × (a + 1)! Which gives (a + 1) = 6 and a = 5 As a result n = 10 But if n is odd, then number of arrangements = a! (a + 1)! Where, n = 2a + 1 When one student is included, number of arrangements = 2 (a + 1)! (a + 1)! Hence, by the GIVEN condition, 2(a + 1) = 3 which is not possible. |
|
| 3. |
1). 482). 573). 504). 120 |
|
Answer» The vowels in the word ‘MENDING’ are I and E. The letters I and E MUST always STAY together while MNDG stay together The letters I and E can be arranged in 2! Ways which is 2 Number of ways, the Vowels and MNDG can be arranged = 6!/2! = 360 Total number of ways such that all the vowels are always together = 360 × 2 = 720 ∴ the letters in the word can be arranged in 720 ways |
|
| 4. |
In how many different ways can the letters of the word “DETAIL” be arranged in such a way that the vowels occupy only the odd positions?1). 602). 363). 324). 28 |
|
Answer» There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants Let the POSITIONS be marked as (1)(2)(3)(4)(5)(6) The 3 vowels can be placed at any of the 3 places marked (1)(3)(5) The number of ways of ARRANGING the vowels = 3P3 = 3! = 6 Also, the 3 consonants can be ARRANGED in the remaining 3 places The number of ways of these arrangements = 3P3 = 3! = 6 TOTAL number of ways = 6 × 6 = 36 ∴ Total number of ways = 36 |
|
| 5. |
Quantity B: Any two digit natural number1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B |
|
Answer» QUANTITY A: Number of WAYS in 3 persons can be selected out of a GROUP of 5 It will be 5C3 = 10 Quantity B: Any two digit natural number will be at LEAST 10. ∴ B ≥ A |
|
| 6. |
The number of 5 digit numbers having their at least one ofdigit repeated is :1). 90002). 697603). 100004). none of these |
|
Answer» No of numbers with REPETITION = 9 X 10 x 10 x 10 x 10 ⇒90000 ( As zero cannot be included at first digit) No.of numbers without repetition = 9 x 9 x 8 x 7 x 6 ⇒ 27216 ∴ No.of numbers with atleast one digit repeated = 90000 − 27216 ⇒ 62784 Hence (4) is the correct answer. |
|
| 7. |
In how many ways can the letters of word SPARROW can be arranged so that the vowels are always at first and last positions?1). 602). 723). 1204). 128 |
|
Answer» There are two vowels, A and O. They can be ARRANGED in first and LAST positions in 2! WAYS. Remaining 5 letters can be arranged in (5!/2!) Ways, as R is repeated. ∴ Number of ways = 2! × (5!/2!) = 120 |
|
| 8. |
In how many ways can 12 different books be divided equally among 4 persons?1). 3000002). 3605003). 3696004). 423500 |
|
Answer» Each person will GET 12 ÷ 4 = 3 books Now, first person can be given 3 books out of 12 DIFFERENT books in 12C3, ways. SECOND person can be given 3 books out of the rest (12 – 3 = 9) books in 9C3 ways. SIMILARLY, third person in 6C3 and the fourth person in 3C3 ways Required no. of ways = 12C3 × 9C3 × 6C3 × 3C3 $(\frac {12!}{3!9!} |
|
| 9. |
1). Quantity B > Quantity A2). Quantity B < Quantity A3). Quantity B ≥ Quantity A4). Quantity B ≤ Quantity A |
|
Answer» Firstly we will find Quantity A, Quantity A: Number of ways to choose 2 students from 5 students = 5C2 = 10 Number of ways to choose 2 TEACHERS from 3 teachers = 3C2 = 3 Number of ways to choose 1 coach from 2 COACHES = 2C1 = 2 ∴ Total number of ways = 10 × 3 × 2 = 60 Now,Quantity B: Total number of PEOPLE = 5 + 3 + 2 = 10 Number of ways to choose 5 people from 10 people = 10C5 = 252 Quantity B > Quantity A
|
|
| 10. |
A committee of five members is to be formed out of 3 Teachers, 4 Assistants and 6 Students. In how many ways can it be done so that the committee either has all 4 Assistants and 1 Student or all 3 Teachers and 2 Assistants?1). 182). 63). 124). 24 |
|
Answer» 4 ASSISTANTS and 1 STUDENT can be selected in 4C4 × 6C1 = 6 ways 3 Teachers and 2 Assistants can be selected in 3C3 × 4C2 = 6 ways ∴ Total ways = 6 + 6 = 12 |
|
| 11. |
In how many ways a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?1). 1202). 2203). 2004). 320 |
|
Answer» Overall we have to select 5 members in all, among these 5, there should be 3 MEN and 2 ladies. Now we have to select 3 men for the committee from the group of 6 men i.e. 6C3 = $(\frac{{6!}}{{3!\LEFT( {6 - 3} \right)!}})$ $(= \frac{{6!}}{{3!.3!}} = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 6}} = 20)$ Similarly, we have to select 2 ladies for the committee from the group of 5 ladies i.e. 5C2 = $(\frac{{5!}}{{2!\left( {5 - 2} \right)!}} = \frac{{5!}}{{2!3!}} = \frac{{120}}{{2 \times 6}} = 10)$ Total NUMBER of POSSIBLE selections= 20 × 10 = 200 |
|
| 12. |
1). 1202). 2403). 2504). 289 |
|
Answer» If digits are all even (0, 2, 4, 6, 8), then we can have 5 × 5 × 5 key values, that is 125 values Similarly, for odd case (1, 3, 5, 7, 9), we have 125 key values ∴ Possible key values are 250(125+125) |
|
| 13. |
In how many ways the letters of the word ‘REFERENCES’ can be arranged so that vowels remain together?1). 25202). 8403). 1104). 105 |
|
Answer» REFERENCES Here? 4 vowels (4 E’s) and 6 consonants (2 R’s, 1 F, 1 N, 1 C and 1 S) ∴ Number of ways to ARRANGE the letters so that vowels are together = 7!/ (4! × 2!) ⇒ 105 |
|