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Permutations and Combinations - Combinations

Grade 11CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Definition of Combination: A combination is a selection of items from a larger pool where the order of selection does not matter. Visually, think of a bag containing 5 different colored balls; if you reach in and grab 2, picking (Red, Blue) is considered the exact same outcome as picking (Blue, Red).

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Difference between Permutations and Combinations: While permutations focus on 'arrangements' where order is key, combinations focus on 'selections' or 'groups'. For example, if 3 students are chosen to stand in a line, their order creates different permutations; however, if those same 3 students are chosen to form a committee, they represent only one single combination regardless of how they stand.

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Complementary Combinations: The number of ways to select rr objects from nn is the same as the number of ways to leave behind n−rn-r objects. This is expressed as nCr=nCn−r^nC_r = ^nC_{n-r}. Visually, if you have a row of 10 light bulbs and you choose 8 to turn 'ON', it is logically the same as choosing 2 specific bulbs to stay 'OFF'.

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Pascal's Rule and the Pascal Triangle: The identity nCr+nCr−1=n+1Cr^nC_r + ^nC_{r-1} = ^{n+1}C_r relates combinations to the construction of Pascal's Triangle. In this triangle, any number (which represents a combination value) is the sum of the two numbers directly above it in the preceding row, illustrating how selections grow as the pool size increases.

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Selection from Distinct Objects: When choosing any number of items from nn distinct objects, each item has two choices: to be included or excluded. This leads to 2n2^n total ways. If at least one item must be selected, we subtract the case where none are chosen, resulting in 2n−12^n - 1 total combinations. Visually, this is like a branching tree where every branch splits into 'Yes' or 'No' for every item in the set.

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Relationship with Permutations: A permutation can be viewed as a two-step process: first, select the items (combination), and then arrange them. Mathematically, nPr=nCr×r!^nP_r = ^nC_r \times r!. Visually, if you select a group of 3 players for a team (nC3^nC_3), and then assign them specific positions like Captain, Vice-Captain, and Keeper (3!3!), you have created permutations.

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Restricted Combinations: Sometimes certain items must always be included or always excluded. If kk particular objects are always included, we only need to choose r−kr-k objects from the remaining n−kn-k objects, resulting in n−kCr−k^{n-k}C_{r-k} ways. Visually, imagine a team selection where the Star Player's spot is already filled; you are effectively choosing from a smaller pool for fewer remaining spots.

📐Formulae

nCr=n!r!(n−r)!, where 0≤r≤n^nC_r = \frac{n!}{r!(n-r)!}, \text{ where } 0 \le r \le n

nC0=nCn=1^nC_0 = ^nC_n = 1

nC1=n^nC_1 = n

nCr=nCn−r^nC_r = ^nC_{n-r}

nCr+nCr−1=n+1Cr^nC_r + ^nC_{r-1} = ^{n+1}C_r

n×n−1Cr−1=(n−r+1)×nCr−1n \times ^{n-1}C_{r-1} = (n-r+1) \times ^nC_{r-1}

If nCx=nCy, then either x=y or x+y=n\text{If } ^nC_x = ^nC_y, \text{ then either } x = y \text{ or } x + y = n

💡Examples

Problem 1:

Find the number of ways to form a committee of 5 members from a group of 6 men and 4 women, such that the committee contains exactly 3 men.

Solution:

  1. We need to select 3 men out of 6 available men. This can be done in 6C3^6C_3 ways.
  2. Since the total committee size is 5, the remaining 5−3=25 - 3 = 2 members must be women. We select 2 women out of 4 available women in 4C2^4C_2 ways.
  3. Using the Multiplication Principle, the total ways = 6C3×4C2^6C_3 \times ^4C_2.
  4. Calculate 6C3=6×5×43×2×1=20^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20.
  5. Calculate 4C2=4×32×1=6^4C_2 = \frac{4 \times 3}{2 \times 1} = 6.
  6. Total ways = 20×6=12020 \times 6 = 120.

Explanation:

This problem uses the basic combination formula to select subsets from two different groups (men and women) and combines them using the fundamental principle of counting.

Problem 2:

If 2nC3:nC3=11:1^{2n}C_3 : ^nC_3 = 11 : 1, find the value of nn.

Solution:

  1. Write the ratio as a fraction: 2nC3nC3=111\frac{^{2n}C_3}{^nC_3} = \frac{11}{1}.
  2. Expand using the formula: (2n)!3!(2n−3)!n!3!(n−3)!=11\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11.
  3. Simplify the expression: (2n)(2n−1)(2n−2)n(n−1)(n−2)=11\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11.
  4. Take common factors: 2n(2n−1)2(n−1)n(n−1)(n−2)=11\frac{2n(2n-1)2(n-1)}{n(n-1)(n-2)} = 11.
  5. Cancel terms nn and (n−1)(n-1): 4(2n−1)n−2=11\frac{4(2n-1)}{n-2} = 11.
  6. Solve for nn: 8n−4=11n−228n - 4 = 11n - 22.
  7. 3n=18  ⟹  n=63n = 18 \implies n = 6.

Explanation:

This example demonstrates how to solve algebraic equations involving combination notations by expanding the factorials and simplifying common terms.