Scenario: You are setting a 4-digit PIN code for your phone using digits 0-9. Case A: Repetition is allowed (e.g., 1122).Case B: Repetition is NOT allowed (e.g., 1234).How many more codes are possible in Case A than in Case B?

Answer: 4960. Slot Method: 4 Slots [ _ _ _ _ ]Case A (Repetition Allowed):$10 \times 10 \times 10 \times 10 = 10,000$ codes.Case B (No Repetition):$10 \times 9 \times 8 \times 7 = 5,040$ codes.(First slot has 10 choices, next has 9...)Difference: $10,000 - 5,040 = 4,960$ ways.

Logic Check: In a cricket team of 11 players, the captain wants to verify two things:1. Selecting 2 openers.2. Deciding who faces the first ball and who faces the second.Which concept applies to each?

Answer: 1: Combination, 2: Permutation. Order Test:1. Selecting Openers: (Rohit, Gill) is the same pair as (Gill, Rohit). Order doesn't matter. $\to$ Combination.2. Striking Strategy: (Rohit on strike, Gill non-strike) is DIFFERENT from (Gill on strike, Rohit non-strike). Order matters. $\to$ Permutation.

Scenario: You are ordering a pizza. The menu offers 3 crust types and 8 toppings.You must choose 1 crust AND exactly 2 different toppings.How many different pizzas can you build?

Answer: 84. Step 1: Choose Crust. 3 options ($^{3}C_{1}$).Step 2: Choose Toppings. Select 2 from 8. Order doesn't matter (Pepperoni, Mushroom is same as Mushroom, Pepperoni). $\to$ $^{8}C_{2}$.Step 3: AND Rule.Total = $^{3}C_{1} \times ^{8}C_{2} = 3 \times \frac{8 \times 7}{2} = 3 \times 28 = 84$.

Visual Logic: 5 friends want to sit on a bench for a picture, but A and B are enemies and refuse to sit next to each other.How many arrangements are possible?

Answer: 72. Subtraction Method (Total - Unwanted):Total ways (without restriction) = $5! = 120$.Unwanted (A & B Together): Tie (AB) as 1 unit. Units: (AB), C, D, E = 4 units $\to 4!$. Internal (A,B) swap = $2!$.Together = $24 \times 2 = 48$.Desired (Not Together): $120 - 48 = 72$.

Identify the Pattern: From a deck of 52 cards, you draw 5 cards. You want to know how many hands contain exactly 3 Kings.What is the correct calculation structure?

Answer: $^{4}C_{3} \times ^{48}C_{2}$. Selection Strategy:1. Choose 3 Kings from 4 available Kings: $^{4}C_{3}$.2. You need 5 cards total, so choose remaining 2 cards from remaining 48 non-King cards: $^{48}C_{2}$.3. AND Rule: Multiply them.Correct: $^{4}C_{3} \times ^{48}C_{2}$.

Usually $^{n}P_{r}$ is used for Permutation and $^{n}C_{r}$ for Combination. What is the mathematical relationship between them?

Answer: $^{n}P_{r} = ^{n}C_{r} \times r!$. $^{n}P_{r}$ (Arrangement) is simply Selection ($^{n}C_{r}$) followed by Arrangement of the selected items ($r!$). Hence, $^{n}P_{r} = ^{n}C_{r} \times r!$.

If you have 5 flags of different colors, how many valid signals can be generated by arranging **at least 2 flags** on a vertical pole?

Answer: 320. Signals can use 2, 3, 4, or 5 flags.2 flags: $5 \times 4 = 20$3 flags: $5 \times 4 \times 3 = 60$4 flags: $5 \times 4 \times 3 \times 2 = 120$5 flags: $5 \times 4 \times 3 \times 2 \times 1 = 120$Total: $20 + 60 + 120 + 120 = 320$.

Why is 0! (factorial zero) equal to 1?

Answer: All of the above.. Mathematically, the empty set can be ordered in exactly one way (the empty order). Also, $1! = 1$, so applying reverse formula $0! = 1!/1 = 1$.

Which grows faster for large values of n?

Answer: Permutation (nPr). Permutation counts arrangements, which multiplies combinations by $r!$. For any $r > 1$, $nPr$ is significantly larger than $nCr$.

In a circular arrangement of 5 people, if clockwise and anti-clockwise orders are treated as DIFFERENT, the total ways are:

Answer: 24. For distinct people where direction matters (standard table seating), formula is $(n-1)!$. Here $(5-1)! = 24$. Only for necklaces/garlands do we divide by 2.

How many words can be formed using letters of the word 'DAUGHTER' so that all vowels come together?

Answer: 4320. Grouping Method:Word: D, A, U, G, H, T, E, R (8 letters).Vowels: A, U, E (Tie them in a bag [AUE]).Consonants: D, G, H, T, R (5 letters).Units: D, G, H, T, R, [AUE] = 6 units $\to 6! = 720$.Internal: [AUE] arranges in $3! = 6$ ways.Total = $720 \times 6 = 4320$.

In how many ways can a team of 5 be selected from 6 men and 5 women such that there are at least 3 men in the team?

Answer: 281. Cases (Men $\ge$ 3):1. 3 Men, 2 Women: $^6C_3 \times ^5C_2 = 20 \times 10 = 200$.2. 4 Men, 1 Woman: $^6C_4 \times ^5C_1 = 15 \times 5 = 75$.3. 5 Men, 0 Women: $^6C_5 \times ^5C_0 = 6 \times 1 = 6$.Total (OR Rule): $200 + 75 + 6 = 281$.

How many 3-digit numbers divisible by 5 can be formed using digits 0, 1, 2, 5, 8 without repetition?

Answer: 21. Numbers divisible by 5 end in 0 or 5.Case 1: Ends with 0: [ _ _ 0 ]First slot: 4 choices (1,2,5,8). Second: 3 choices. $\to 4 \times 3 \times 1 = 12$.Case 2: Ends with 5: [ _ _ 5 ]First slot: 3 choices (can't be 0, can't be 5). Digits avail: 1,2,8. Second slot: 3 choices (0 allowed now). $\to 3 \times 3 \times 1 = 9$.Total: $12 + 9 = 21$.

In how many ways can 6 people sit around a circular table if X and Y must always sit together?

Answer: 48. Grouping + Circular:1. Group (XY) as 1 unit. Remaining 4 people. Total 5 units.2. Arrange 5 units in circle = $(5-1)! = 4! = 24$.3. Internal arrangement of X, Y = $2! = 2$.Total = $24 \times 2 = 48$.

How many diagonals are there in a decagon (10-sided polygon)?

Answer: 35. Diagonal Formula: $\frac{n(n-3)}{2}$For Decagon, $n=10$.$\text{Diagonals} = \frac{10(10-3)}{2} = \frac{10 \times 7}{2} = 35$.Logic: $^{10}C_2$ (lines) - 10 (sides) = $45 - 10 = 35$.

In how many ways can the letters of the word 'BANANA' be arranged?

Answer: 60. Total letters = 6. Repeats: A (3 times), N (2 times).Formula: $\frac{6!}{3! \times 2!} = \frac{720}{6 \times 2} = \frac{720}{12} = 60$.

How many sub-committees of 3 people can be formed from a committee of 7 people?

Answer: 35. Order does not matter (A,B,C is same committee as B,C,A). Use Combination.$^{7}C_{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$.

If $^{15}P_{r} = 2730$, find the value of r.

Answer: 3. Start multiplying downwards from 15: $15 \times 14 = 210$.$210 \times 13 = 2730$.We multiplied 3 terms (15, 14, 13). So, $r = 3$.

4 boys and 4 girls sit in a row. How many ways can they sit if they must sit **alternately**?

Answer: 1152. Case 1: B G B G B G B G (Start with Boy) $\to$ Boys: $4!$, Girls: $4!$. Total: $24 \times 24 = 576$.Case 2: G B G B G B G B (Start with Girl) $\to$ Same: $576$.Total: $576 + 576 = 1152$.

A polygon has 54 diagonals. Number of sides of this polygon is:

Answer: 12. Formula: $\frac{n(n-3)}{2} = 54$.$n(n-3) = 108$.Think of two factors of 108 with gap 3. $12 \times 9 = 108$. So, $n=12$.

Challenger: 4 letters are to be put into 4 addressed envelopes. In how many ways can they be put such that exactly one letter goes into the wrong envelope?

Answer: 0. Logic Trap: It is IMPOSSIBLE for exactly one letter to be wrong. If letter A goes to B's envelope, then letter B must also go somewhere wrong. Misplacements always happen in pairs or cycles. So, the answer is 0.

What is the rank of the word 'SACHIN' in the dictionary?

Answer: 601. Alphabetical Order: A, C, H, I, N, S.1. Fix A_ _ _ _ _ $\to 5! = 120$ words.2. Fix C_ _ _ _ _ $\to 120$ words.3. Fix H_ _ _ _ _ $\to 120$ words.4. Fix I_ _ _ _ _ $\to 120$ words.5. Fix N_ _ _ _ _ $\to 120$ words.Total so far = 600. Next word starts with S, then A,C,H,I,N (Matches SACHIN).Rank = 600 + 1 = 601.

A box contains 5 red, 4 green, and 3 blue balls. Three balls are drawn. What is the probability that at least one is Blue? (Use Selection Logic)

Answer: 34/55. Total Balls = 12. Total ways to select 3 = $^{12}C_{3} = 220$.Calculate Inverse: No Blue ball drawn. Draw 3 from (5R+4G=9). $^{9}C_{3} = 84$.Ways with at least 1 Blue = Total - No Blue = 220 - 84 = 136.Prob = 136/220 = 34/55.

There are 10 people in a room. If everyone shakes hands with everyone else exactly once, how many handshakes take place?

Answer: $45$. Selection Logic: A handshake requires 2 people. Order doesn't matter (A shaking B's hand is same as B shaking A's).Formula: $^{10}C_{2} = \frac{10 \times 9}{2} = 45$.

How many triangles can be formed by joining 10 points on a plane, out of which 4 are collinear?

Answer: $116$. Total - Impossible:Total triangles from 10 points: $^{10}C_{3} = \frac{10 \times 9 \times 8}{6} = 120$.Impossible triangles from 4 collinear points: $^{4}C_{3} = 4$.Valid Triangles = $120 - 4 = 116$.

Find the sum of all 4-digit numbers formed by using the digits 1, 2, 3, 4 (without repetition).

Answer: $66660$. Shortcut Formula: Sum = (Sum of digits) $\times (n-1)! \times (1111... n \text{ times})$.Sum = $(1+2+3+4) \times 3! \times 1111$.Sum = $10 \times 6 \times 1111 = 60 \times 1111 = 66,660$.

In how many ways can 6 different beads be strung into a necklace?

Answer: $60$. Necklace Rule: For a necklace, clockwise and anti-clockwise are same (flip).Formula: $\frac{(n-1)!}{2}$.$n=6 \to \frac{5!}{2} = \frac{120}{2} = 60$.

Grid Problem: A person moves from (0,0) to (4,3) on a grid only moving RIGHT (R) or UP (U). How many distinct paths exist?

Answer: $35$. Total Steps: To go from (0,0) to (4,3), you need 4 Right moves and 3 Up moves. Total steps = 7.Path is a permutation of RRRRUUU.Formula: $\frac{7!}{4! \times 3!} = \frac{5040}{24 \times 6} = 35$.Alternatively: $^{7}C_{3} = 35$.

If a set $A = \{1, 2, 3, 4, 5\}$, find the number of subsets of A that contain the element '2' but NOT '5'.

Answer: $8$. Constraint Logic:- Must include '2': One way.- Must NOT include '5': One way (exclude).- Remaining elements $\{1, 3, 4\}$: Each can be in or out (2 choices each).Total = $2 \times 2 \times 2 = 2^3 = 8$.

A box contains 2 White, 3 Black, and 4 Red balls. In how many ways can 3 balls be drawn such that at least one is Black?

Answer: $64$. Total - None:Total balls = 9. Total ways = $^{9}C_{3} = 84$.No Black (Draw from 2W + 4R = 6 balls) = $^{6}C_{3} = 20$.At least 1 Black = $84 - 20 = 64$.

Usually $^{n}P_{r}$ is used for Permutation and $^{n}C_{r}$ for Combination. What is the mathematical relationship between them?

$^{n}P_{r} = ^{n}C_{r} \times r!$

$^{n}C_{r} = ^{n}P_{r} \times r!$

Why is 0! (factorial zero) equal to 1?

It is just a defined rule by mathematicians.

Number of ways to arrange 0 items is 1 (do nothing).

It completes the pattern n! = (n+1)! / (n+1).

Home/Rajasthan 4th Grade Exam/Quantitative Aptitude/

Permutation & Combination

Master the art of counting without counting. Learn the Slot Method, Grouping Tricks, and Rank Shortcuts to solve complex arrangement and selection problems with zero confusion.

Expert Answer & Key Takeaways

Master the art of counting without counting. Learn the Slot Method, Grouping Tricks, and Rank Shortcuts to solve complex arrangement and selection problems with zero confusion.

1. The Fundamental Counting Principle

Before formulas, understand logic.

The AND Rule (Multiplication): If you must do Task A AND Task B, multiply their ways.
Example: 3 Shirts AND 2 Pants =

3 \times 2 = 6

outfits.

The OR Rule (Addition): If you must do Task A OR Task B, add their ways.
Example: Go by Bus (3 types) OR Train (2 types) =

3 + 2 = 5

ways.

Example:

Q: A room has 2 doors and 3 windows. In how many ways can a thief enter through a door and exit through a window?

Solution: Enter (2 ways) AND Exit (3 ways) =

2 \times 3 = 6

ways.

2. Permutation vs. Combination: The 'Order Test'

The biggest confusion is 'Which formula to use?'. Apply the Order Test:

1. Change the order of selected items.
2. Does it make a new outcome?
- YES

\to

Order Matters

\to

Permutation (Arrangement)
- NO

\to

Order Doesn't Matter

\to

Combination (Selection)

Keywords:
- Permutation: Arrange, Line up, Code, Rank, Position.
- Combination: Select, Choose, Team, Group, Handshake.

Example:

Q: Selecting 3 students for a photo vs. Selecting 3 students for a trip.

Solution: Photo: Order matters (Left/Right position changes photo)

\to

Permutation.
Trip: Order doesn't matter (Ram, Shyam, Geeta is same group as Geeta, Ram, Shyam)

\to

Combination.

3. The 'Slot Method' (Sniper Method)

Forget

nPr

! Use the Slot Method for arrangements.

Step 1: Draw empty slots (boxes) equal to the number of positions.
Step 2: Fill each slot with the number of options available for that position.
Step 3: Multiply them.

Visual Model:
Arranging 3 people on 5 chairs?
[5 options]

\times

[4 options]

\times

[3 options] = 60 ways.
(First person has 5 choices, next has 4, next has 3).

Example:

Q: How many 3-digit numbers can be formed using 1, 2, 3, 4, 5 without repetition?

Solution: 3 Slots: [ _ _ _ ]
1st Slot: 5 choices. 2nd Slot: 4 choices. 3rd Slot: 3 choices.
Total =

5 \times 4 \times 3 = 60

4. Grouping Method (String Method)

Use this when items must 'always keep together'.

The Trick:
1. Tie the 'together' items with a string and treat them as ONE unit.
2. Arrange the main units.
3. Multiply by the internal arrangement of the tied group.

Example:

Q: Arrange 'APPLE' so P's are together.

Solution: Units: [A], [L], [E], [PP]. Total 4 units

\to

4!

Inside [PP]:

\frac{2!}{2!} = 1

way.
Total =

4! \times 1 = 24

5. Gap Method (Items Never Together)

Use this when items must 'never sit together'.

The Trick:
1. Arrange the 'others' first (creates gaps).
2. Place the specific items in the Gaps (spaces between adjacent items + ends).

Visual:
_ O _ O _ O _ (3 Others create 4 Gaps).
Select gaps using

nCr

, then arrange.

6. Circular Permutation Hack

In a circle, there is no 'start' or 'end'.

Rule 1 (People/Distinct Items):

(n-1)!

- Fix one person to create a reference point, then arrange remaining

(n-1)

.

Rule 2 (Necklace/Garland/Identical Flip):

\frac{(n-1)!}{2}

- Clockwise and Anti-clockwise look same (flippable).

7. Dictionary Rank Shortcut

Find rank of word 'MAKE' in dictionary.
1. Write letters alphabetically: A, E, K, M.
2. Fix 'A' first

\to

remaining 3! words. (Does target match? No, target starts with M).
3. Fix 'E'

\to

3! words.
4. Fix 'K'

\to

3! words.
5. Fix 'M'

\to

Match! Now fix next letter 'A'...

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Permutation & Combination

Expert Answer & Key Takeaways

1. The Fundamental Counting Principle

2. Permutation vs. Combination: The 'Order Test'

3. The 'Slot Method' (Sniper Method)

4. Grouping Method (String Method)

5. Gap Method (Items Never Together)

6. Circular Permutation Hack

7. Dictionary Rank Shortcut

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