Question 1

Form 5-digit numbers without repetition using digits {0, 1, 2, 3, 4, 5}. How many such numbers are divisible by 3? (The first digit cannot be 0.)

Accepted Answer

Introduction / Context: A number is divisible by 3 if the sum of its digits is a multiple of 3. We must count 5-digit, no-repetition numbers from the 6 digits {0,1,2,3,4,5} with a non-zero leading digit. Given Data / Assumptions: Available digits: 0–5. Distinct digits; 5-digit numbers; first digit ≠ 0. Sum of all six digits is 15, divisible by 3. Concept / Approach: Select which single digit is omitted; the remaining 5 digits define the digit-sum. If the total sum of chosen digits is divisible by 3, every permutation yields a number divisible by 3 (subject to the no-leading-zero rule). Step-by-Step Solution: Let S6 = 0+1+2+3+4+5 = 15 ≡ 0 (mod 3).Sum of chosen five digits ≡ −(omitted digit) (mod 3). To get 0 (mod 3), the omitted digit must be 0 or 3.Case 1 (omit 0): digits {1,2,3,4,5} ⇒ all 5! = 120 arrangements valid.Case 2 (omit 3): digits {0,1,2,4,5} ⇒ total 5! = 120 permutations; subtract leading-zero cases (0 as first): 4! = 24 ⇒ 96 valid.Total = 120 + 96 = 216. Verification / Alternative check: Residue-class reasoning (mod 3) matches the omitted-digit argument; direct enumeration agrees. Why Other Options Are Wrong: 210/217 are near but incorrect; 122/192 do not follow from the modular analysis. Common Pitfalls: Forgetting to exclude leading-zero permutations in the case where 0 is present. Final Answer: 216

Question 2

A new flag has six vertical stripes, each stripe colored using some or all of the colors {yellow, green, blue, red}. No two adjacent stripes may have the same color. In how many distinct stripe colorings can the flag be made?

Accepted Answer

Introduction / Context: This counts sequences of length 6 over a 4-color alphabet with an adjacency restriction that prohibits identical neighbors. Given Data / Assumptions: Colors available independently per stripe: 4 choices initially. No two consecutive stripes may have the same color. Concept / Approach: Use multiplication with a recurrence flavor: first stripe has 4 choices; each subsequent stripe has 3 choices (anything but its immediate neighbor’s color). Step-by-Step Solution: Stripe 1: 4 ways.Stripes 2–6: each 3 ways independently of earlier stripes beyond the immediate neighbor.Total = 4 × 3^5 = 4 × 243 = 972 = 12 × 81. Verification / Alternative check: A simple recurrence a(n) = 3 a(n−1) with a(1) = 4 yields a(6) = 4 × 3^5. Why Other Options Are Wrong: Other products do not equal 972 and do not match the 4·3^5 logic. Common Pitfalls: Accidentally allowing 4 choices at each step or forbidding more than just the immediate neighbor’s color. Final Answer: 12 x 81

Question 3

An intelligence agency forms 2-digit codes from distinct digits 0–9, with the first digit non-zero. A handwritten code can be misread upside down if both digits are from {0,1,6,8,9} and the 180° rotation maps it to another valid 2-digit code (e.g., …

Accepted Answer

Introduction / Context: Under 180° rotation, the only digits that remain valid digits are {0,1,6,8,9} with mappings 0→0, 1→1, 6↔9, 8→8. A code is confusable if both digits come from this set and its rotated reading is a different valid 2-digit code (non-zero first digit, digits distinct). Given Data / Assumptions: Total 2-digit codes with distinct digits and non-zero first digit: 9 × 9 = 81. Rotation map f: {0,1,6,8,9} → {0,1,9,8,6} and order reverses. Concept / Approach: Count confusable codes and subtract from 81. Confusion requires both digits from {1,6,8,9} (not 0 for the second digit), so that the rotated code also has a non-zero leading digit and remains a valid 2-digit code with distinct digits. Step-by-Step Solution: Confusable codes = ordered pairs (x, y) with x ∈ {1,6,8,9}, y ∈ {1,6,8,9}, x ≠ y, because the rotated code is f(y)f(x) with f(y) ∈ {1,6,8,9} ≠ 0 and f(x) ≠ f(y).Count = 4 × 3 = 12.Therefore, non-confusable codes = 81 − 12 = 69. Verification / Alternative check: Cases with y = 0 yield rotated codes starting with 0 and hence are not considered confusing under the “valid code” criterion. Why Other Options Are Wrong: 80/78 assume excluding very few cases; 71 is close but overcounts confusable cases; 81 ignores the confusion constraint. Common Pitfalls: Treating 0 as allowable as the rotated leading digit; or allowing equal digits (not permitted). Final Answer: 69

Question 4

Using the digits {1, 2, 3, 4, 5} without repetition, form 5-digit numbers. How many such numbers have the units (ones) digit greater than the tens digit?

Accepted Answer

Introduction / Context: We count permutations with a simple inequality constraint involving only the last two positions (tens and units digits). Given Data / Assumptions: Digits are 1–5, used once each. 5-digit permutations (all digits used). Constraint: units digit > tens digit. Concept / Approach: Symmetry on ordered pairs: for any unordered pair {a, b} with a ≠ b, exactly one of the two orders satisfies units > tens. Step-by-Step Solution: Choose which two digits occupy the last two places: C(5, 2) = 10 unordered choices; for each chosen pair, exactly 1 ordering has units > tens.Arrange the first three positions with the remaining 3 digits: 3! = 6.Total = 10 × 6 = 60. Verification / Alternative check: Out of all 5P2 = 20 ordered pairs for the last two positions, half satisfy the inequality; 20/2 = 10 ways for the last two, times 6 ways for the first three gives 60. Why Other Options Are Wrong: 48 (= 2 × 4!) ignores the exact half-ordering logic; 54/72 are ad hoc counts. Common Pitfalls: Double-counting when choosing ordered pairs first; working with unordered pairs avoids that. Final Answer: 60

Question 5

Five boxes in a row (positions 1–5) are each filled with exactly one ball, either red (R) or blue (B). No two adjacent boxes may both contain blue balls. Balls of the same color are identical. How many different arrangements (strings over {R, B}) ar…

Accepted Answer

Introduction / Context: We are counting binary strings of length 5 with no consecutive B’s. This classic constraint yields a Fibonacci-type recurrence. Given Data / Assumptions: Length n = 5. Alphabet {R, B} with “no BB” adjacency. R and B are indistinguishable beyond color (identical balls). Concept / Approach: Let a(n) be the number of valid strings of length n. Appending R to any valid string of length n−1 is always allowed; appending B is allowed only to strings of length n−1 that end in R. This leads to a(n) = a(n−1) + a(n−2), a Fibonacci recurrence. Step-by-Step Solution: Base cases: a(1) = 2 (R, B); a(2) = 3 (RR, RB, BR).Then a(3) = 5; a(4) = 8; a(5) = 13.Therefore, there are 13 valid arrangements for 5 boxes. Verification / Alternative check: Direct enumeration confirms: 8 with at most two B’s spaced apart plus 5 with exactly two B’s non-adjacent. Why Other Options Are Wrong: 8 and 10 undercount; 15 and 22 overcount by allowing adjacent B’s. Common Pitfalls: Assuming each position always has 2 independent choices (which ignores the adjacency ban) or forgetting that RB and BR are both valid. Final Answer: 13

Question 6

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Accepted Answer

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = ( 7 C 3 * 4 C 2 ) = 210. Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. Number of ways of arranging 5 letters among themselves = 5! = 120 Required number of ways = (210 x 120) = 25200.

Question 7

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

Accepted Answer

The smallest number in the series is 1000, a 4-digit number. The largest number in the series is 4000, the only 4-digit number to start with 4. The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3. The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4. Hence, there are 3 x 5 x 5 x 5 or 375 numbers from 1000 to 3999. Including 4000, there will be 376 such numbers.

Question 8

A committee of 5 persons is to be formed from 6 men and 4 women. In how many ways can this be done when at least 2 women are included ?

Accepted Answer

When at least 2 women are included. The committee may consist of 3 women, 2 men : It can be done in 4 C 3 * 6 C 2 ways or, 4 women, 1 man : It can be done in 4 C 4 * 6 C 1 ways or, 2 women, 3 men : It can be done in 4 C 2 * 6 C 3 ways. Total number of ways of forming the committees = 4 C 2 * 6 C 3 + 4 C 3 * 6 C 2 + 4 C 4 * 6 C 1 = 6 x 20 + 4 x 15 + 1x 6 = 120 + 60 + 6 =186

Question 9

If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word ‘SACHIN’ appears at serial number :

Accepted Answer

If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways. If it started with c then the remaining 5 positions can be filled in 5! Ways.Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways. If it started with S then the remaining position can be filled with A,C,H,I,N in alphabetical order as on dictionary. The required word SACHIN can be obtained after the 5X5!=600 Ways i.e. SACHIN is the 601th letter.

Question 10

12 people at a party shake hands once with everyone else in the room.How many handshakes took place?

Accepted Answer

There are 12 people, so this is our n value. So, 12 C 2 1= 66

Question 11

In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?

Accepted Answer

The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Question 12

How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane?

Accepted Answer

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines. AB , AC BA , BC CA , CB There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB. The lines are: AB, BC and AC ; 3 lines only. So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

Question 13

In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?

Accepted Answer

The word 'OPTICAL' contains 7 different letters. When the vowels OIA are always together, they can be supposed to form one letter. Then, we have to arrange the letters PTCL (OIA). Now, 5 letters can be arranged in 5! = 120 ways. The vowels (OIA) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.

Question 14

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

Accepted Answer

Required number of ways = ( 7 C 5 * 3 C 2 ) = ( 7 C 2 * 3 C 1 ) = 63

Question 15

A college has 10 basketball players. A 5-member team and a captain will be selected out of these 10 players. How many different selections can be made?

Accepted Answer

A team of 6 members has to be selected from the 10 players. This can be done in 10 C 6 or 210 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210×6= 1260

Question 16

How many arrangements can be made out of the letters of the word COMMITTEE, taken all at a time, such that the four vowels do not come together?

Accepted Answer

There are total 9 letters in the word COMMITTEE in which there are 2M's, 2T's, 2E's. The number of ways in which 9 letters can be arranged = 9 ! 2 ! × 2 ! × 2 ! = 45360 There are 4 vowels O,I,E,E in the given word. If the four vowels always come together, taking them as one letter we have to arrange 5 + 1 = 6 letters which include 2Ms and 2Ts and this be done in 6 ! 2 ! × 2 ! = 180 ways. In which of 180 ways, the 4 vowels O,I,E,E remaining together can be arranged in 4 ! 2 ! = 12 ways. The number of ways in which the four vowels always come together = 180 x 12 = 2160. Hence, the required number of ways in which the four vowels do not come together = 45360 - 2160 = 43200

Question 17

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

Accepted Answer

'LOGARITHMS' contains 10 different letters. Required number of words = Number of arrangements of 10 letters, taking 4 at a time. = 10 P 4 = 5040.

Question 18

When four fair dice are rolled simultaneously, in how many outcomes will at least one of the dice show 3?

Accepted Answer

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes. The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes. Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

Question 19

In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of p…

Accepted Answer

The number of points of intersection of 37 lines is C 2 37 . But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting C 2 13 points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting C 2 11 points, we get only one point B. Hence the number of intersection points of the lines is C 2 37 - C 2 13 - C 2 11 + 2 = 535

Question 20

How many necklace of 12 beads each can be made from 18 beads of different colours?

Accepted Answer

Here clock-wise and anti-clockwise arrangements are same. Hence total number of circular–permutations: 18 P 12 2 * 12 = 18 ! 6 * 24

Permutation and Combination Questions