COMBINATION WORD PROBLEMS EXAMPLES

A combination is a selection of objects without regard to order.

Problem 1 :

How many different teams of 4 can be selected from a squad of 7 if,

i) there are no restrictions

ii) the teams must include the captain ?

Solution :

Total number of players = 7

number of players to be selected = 4

i) since there is no restriction, the number of ways be 7C4

= 7!/(7-4)! 4!

ii) There must be 1 player to be selected for captain. From the remaining 6 player, 4 players should be selected.

Problem 2 :

If there are 12 persons in a party and if each two of them shake hands with each other, how many handshakes happen in the party ?

Solution :

12C2 = 12!(12-2)! 2!= 12!10! 2!= 12×11×10!10! 2!= 6×11= 66

Problem 3 :

A question paper has two parts, Part A and Part B each containing 10 questions. If a student to choose 8 from Part A and 5 from Part B, in how many ways can he choose the questions ?

Solution :

  • Part A is containing 10 questions, from those 10 questions 8 questions should be selected.
  • Part B is containing 10 questions, from those 10 questions 5 questions should be selected.
10C8×10C5 = 10!(10-8)! 8!10!(10-5)! 5!= 10×9×8!2! 8!10×9×8×7×6×5!5! 5!

So, the required number of ways are 11340.

Problem 4 :

In how many ways a committee of 5 members can be selected from 6 men and 5 women consisting of 3 men and 2 women ?

Solution :

6C3×5C2 = 6!(6-3)! 3!×5!(5-2)! 2!= 6!3! 3!×5!3! 2!= 6×5×4×3!3×2×1! 3!×5×4×3!3! 2×1= 5×4×5×2= 200 ways

Problem 5 :

In how many ways can a cricket eleven be chosen out of a batch of 15 players if

i) there is no restriction on the selection

ii)  a particular players is always chosen ?

iii)  a particular player is never chosen ?

Solution :

i)  Total number of players = 15

number of players to be selected = 11

15C11= 15!(15-11)! 11!= 15×14×13×12×11!4! 11!= 1365 ways

ii)  one person always to be chosen means, 10 persons are selected from the total of 14 players.

14C10= 14!(14-10)! 10!= 14×13×12×11×10!4! 10!= 1001 ways

iii) By leaving one particular person, we have total of 14 choices.

14C11= 14!(14-11)! 11!= 14×13×12×11!3! 11!= 364 ways

Problem 6 :

How many chords can be drawn through 21 points on a circle?

Solution :

By joining any two points on the circle, we can create a chord. Even diameter is also chord, but chord is not a diameter.

21C2= 21!(21-2)! 2!= 21×20×19!19! 2!= 210

Problem 7 :

From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide either all of them will join or none of them will join. In how many ways can they be chosen ?

Solution :

i) All 3 students should join :

Let us consider the three students has been joined and the remaining 7 students alone can be selected from 22.

22C7= 22!(22-7)! 7!= 22×21×20×19×18×17×16×15!15! 7!=

ii) Those three students doesn't join :

22C10= 22!(22-10)! 10!= 22×21×20×19×18×17×16×15×14×13×12!12! 10!=

Number of ways that any one of it happen :

(1) + (2)

= 170544 + 646646

= 817190

Recent Articles

  1. Finding Range of Values Inequality Problems

    May 21, 24 08:51 PM

    Finding Range of Values Inequality Problems

    Read More

  2. Solving Two Step Inequality Word Problems

    May 21, 24 08:51 AM

    Solving Two Step Inequality Word Problems

    Read More

  3. Exponential Function Context and Data Modeling

    May 20, 24 10:45 PM

    Exponential Function Context and Data Modeling

    Read More