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Combination Concepts
QUANTITATIVEAPTITUDE

Combination Concepts

Learn combinations, selection formulas, and counting techniques for aptitude problems.

1. Basic Combination / Selection

  • What is the question? These questions ask:

    “How many ways can we select rr people/items from nn people/items?”

    Order does not matter.

  • Formula:

    nCr=n!r!(n−r)!^nC_r=\frac{n!}{r!(n-r)!}

  • Example: From 10 students, how many ways can we select 3 students?

    10C3=10!3!7!^{10}C_3=\frac{10!}{3!7!}

    =10×9×83×2×1=\frac{10\times9\times8}{3\times2\times1}

    =120=120

    Answer: 120 ways

2. Selection with Mandatory Members

  • What is the question? A certain person or group must be selected.

  • Formula:

    Required selections=n−kCr−k\text{Required selections}={}^{n-k}C_{r-k}

    where kk people are already fixed.

  • Example: A committee of 5 is selected from 10 people. Two specific people must be included. How many committees are possible?

    The 2 people are already selected.

    Total members needed = 5
Already selected      = 2
Still needed          = 3
People available      = 10 - 2 = 8

    Therefore:

    8C3=56^8C_3=56

    Answer: 56 ways

3. Selection with Excluded Members

  • What is the question? One or more particular people cannot be selected.

  • Formula:

    Ways=n−kCr\text{Ways}={}^{n-k}C_r

    where kk people are excluded.

  • Example: A committee of 4 is selected from 10 people. Two particular people cannot be included. How many committees are possible?

    Available people = 10 - 2 = 8

    Choose 4:

    8C4=70^8C_4=70

    Answer: 70 ways

4. Selection with “At Least”

  • What is the question? The question says:

    “Select a team with at least 2 women.”

    “At least” means 2 or more.

  • Formula:

    Break the problem into cases and add them.

  • Example: A team of 6 is selected from 8 men and 4 women. How many teams contain at least 2 women?

    Possible cases:

    2 women + 4 men
3 women + 3 men
4 women + 2 men

    Therefore:

    +{}^4C_3\times{}^8C_3 +{}^4C_4\times{}^8C_2$$ $$=6(70)+4(56)+1(28)$$ $$=420+224+28$$ $$=672$$ **Answer: 672 ways**

5. Selection with “At Most”

  • What is the question? The question says:

    “Select a team with at most 2 women.”

    “At most” means 2 or fewer.

  • Formula:

    Add the cases from 00 up to the given number.

  • Example: A team of 4 is selected from 5 men and 3 women. How many teams contain at most 2 women?

    Possible cases:

    0 women + 4 men
1 woman  + 3 men
2 women  + 2 men

    Therefore:

    +{}^3C_1{}^5C_3 +{}^3C_2{}^5C_2$$ $$=1(5)+3(10)+3(10)$$ $$=5+30+30$$ $$=65$$ **Answer: 65 ways**

6. Selection of People from Different Groups

  • What is the question? The question gives a fixed number to be selected from each group.

    For example:

    “Select 2 men and 3 women.”

  • Formula:

    aCx×bCy^aC_x\times{}^bC_y

  • Example: A team must contain 2 men from 6 men and 3 women from 5 women. How many teams are possible?

    Select the men:

    6C2=15^6C_2=15

    Select the women:

    5C3=10^5C_3=10

    Therefore:

    15×10=15015\times10=150

    Answer: 150 ways

7. Selection with Exactly

  • What is the question? The question says:

    “Select exactly 2 women.”

    “Exactly” means only that specific number.

  • Formula:

    wCk×mCr−k^wC_k\times{}^mC_{r-k}

  • Example: A team of 5 is selected from 6 men and 4 women. How many teams contain exactly 2 women?

    Select 2 women:

    4C2=6^4C_2=6

    Select remaining 3 men:

    6C3=20^6C_3=20

    Therefore:

    6×20=1206\times20=120

    Answer: 120 ways

8. Forming Lines from Points

  • What is the question? Find how many different straight lines can be formed using a set of points.

  • Formula:

    nC2^nC_2

    because 2 points determine a line.

  • Example: How many lines can be formed using 8 points, assuming no three points are collinear?

    8C2^8C_2

    =8×72=\frac{8\times7}{2}

    =28=28

    Answer: 28 lines

9. Forming Triangles from Points

  • What is the question? Find how many triangles can be formed from a set of points.

  • Formula:

    nC3^nC_3

    because a triangle requires 3 points.

  • Example: 8 points are placed on a circle. How many triangles can be formed?

    Any 3 points form a triangle:

    8C3=56^8C_3=56

    Answer: 56 triangles

10. Identical Objects Distributed Among People

  • What is the question? Identical objects are distributed among different people, and objects of the same type are indistinguishable.

    Example:

    “Distribute 10 identical chocolates among 3 children.”

  • Formula:

    n+r−1Cr−1^{n+r-1}C_{r-1}

    for distributing nn identical objects among rr distinct people when zero is allowed.

  • Example: Distribute 10 identical chocolates among 3 children. A child may receive zero chocolates.

    10+3−1C3−1^{10+3-1}C_{3-1}

    12C2=66^{12}C_2=66

    Answer: 66 ways

11. Identical Objects with At Least One Each

  • What is the question? Identical objects must be distributed so that every person gets at least one.

  • Formula:

    First give 1 object to each person.

    Remaining=n−r\text{Remaining}=n-r

    Then:

    n−1Cr−1^{n-1}C_{r-1}

  • Example: Distribute 10 identical chocolates among 3 children so that every child gets at least one.

    Give each child 1:

    10 - 3 = 7 chocolates remain

    Distribute the remaining 7 freely:

    7+3−1C3−1^{7+3-1}C_{3-1}

    9C2=36^9C_2=36

    Answer: 36 ways

Advanced Variants

12. Selection with “At Least One” from a Group

  • What is the question? Select a team while making sure that at least one person comes from a particular group.

  • Formula:

    Required ways=Total ways−Ways with none\text{Required ways}=\text{Total ways}-\text{Ways with none}

  • Example: A committee of 4 is selected from 6 men and 3 women. How many committees contain at least one woman?

    Total committees:

    9C4=126^9C_4=126

    Committees with no women:

    6C4=15^6C_4=15

    Therefore:

    126−15=111126-15=111

    Answer: 111 ways

13. Selection with People Sitting Together / Team Conditions

  • What is the question? These questions combine selection and arrangement, such as:

    “Select 4 people from 8 and arrange them in a row.”

  • Formula:

    nCr×r!=nPr^nC_r\times r! = {}^nP_r

  • Example: From 8 people, select and arrange 3 people in a row.

    Select 3:

    8C3=56^8C_3=56

    Arrange them:

    3!=63!=6

    Therefore:

    56×6=33656\times6=336

    Answer: 336 ways

14. Committee with a Pair That Must Stay Together

  • What is the question? Two specific people must either both be selected or both be excluded.

  • Example: A committee of 4 is selected from 8 people. Two specific people must either both be selected or both excluded. Find the number of committees.

    Case 1: Both selected

    Remaining 2 from the other 6:

    6C2=15^6C_2=15

    Case 2: Both excluded

    Select all 4 from the remaining 6:

    6C4=15^6C_4=15

    Total:

    15+15=3015+15=30

    Answer: 30 ways

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