 Mathematics COMBINATIONS - BASICS AND DEFINITION
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### Topics Covered

star Combinations
star Relation between Permutation and Combination formula

### Combinations => Let us now assume that there is a group of 3 lawn tennis players color(blue)(X, Y, Z.) A team consisting of 2 players is to be formed.

color(blue)("In how many ways can we do so ?")

Here, color(blue)("order is not important.") In fact, there are only 3 possible ways in which the team could be constructed.

These are color(blue)(XY, YZ) and color(blue)(ZX) (Fig 1. )

Here, each selection is called a combination of 3 different objects taken 2 at a time. In a combination, the order is not important.

Some more illustrations where order is not important

Twelve persons meet in a room and each shakes hand with all the others. How do we determine the number of hand shakes. X shaking hands with Y and Y with X will not be two different hand shakes. Here, order is not important. There will be as many hand shakes as there are combinations of 12 different things taken 2 at a time.

Seven points lie on a circle. How many chords can be drawn by joining thesepoints pairwise? There will be as many chords as there are combinations of 7 different things taken 2 at a time.

Now, we obtain the formula for finding the number of combinations of n different objects taken r at a time, denoted by color(red)(text()^nC_r).

=> color(green)("Suppose we have 4 different objects A, B, C and D. Taking 2 at a time"),

if we have to make combinations, these will be AB, AC, AD, BC, BD, CD.

Here, AB and BA are the same combination as order does not alter the combination.

This is why we have not included BA, CA, DA, CB, DB and DC in this list.

There are as many as 6 combinations of 4 different objects taken 2 at a time, i.e., text()^4C_2 = 6.

Corresponding to each combination in the list, we can arrive at 2! permutations as 2 objects in each combination can be rearranged in 2! ways. Hence, the number of permutations = text()^4C_2 × 2!.

On the other hand, the number of permutations of 4 different things taken 2 at a time = text()^4P_2.

Therefore text()^4P_2 = text()^4C_2 × 2! or (4!)/((4-2)! 2!)= text()^4C_2
\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT"

color(red)(★ ul"Combinations")

Combination is selection of one or more things out of n things which may be alike or different taken some oral! at a time.

it is denoted by color(red)(text()^nC_r).

### Relation between Permutation and Combination formula

 color(blue)"Theorem - " color(red)(text()^nP_r = text()^nC_r × r! , \ \ \ \ \ 0 < r ≤ n.)

"Proof" Corresponding to each combination of text()^nC_r  we have r ! permutations,

because r objects in every combination can be rearranged in r ! ways.

Hence, the total number of permutations of n different things taken r at a time is text()^nC_r × r! , On the other hand, it is text()^nP_r .

Thus text()^nP_r = text()^nC_r × r! , \ \ \ \ \ 0 < r ≤ n.

\color{green} ★ \color{green} \mathbf(KEY \ POINTS)

\color{green} (1-) From above  (n!)/( ( n-r)!) = text()^(n)C_r xx r!  , i.e., text()^(n)C_r = (n!)/(r! ( n-r)! )

In particular, if r = n , color(red)(text()^nC_n = (n!)/(n! 0!) =1) .

\color{green} (2-) As (n!)/(0! (n-0)! ) =color(red)(1 = text()^(n)C_0), the formula text()^nC_r = (n!)/( (n-r)! r!)  is applicable for r = 0 also.

Hence

color(red)(text()^(n)C_r = (n!)/( (n-r)! r!)) , 0 le r le n .

\color{green} (3-) text()^(n)C_(n-r) = (n!)/( (n-r)! ( n - ( n-r) )! ) = (n!)/( (n-r)! r!) = text()^(n)C_r .,

i.e., selecting r objects out of n objects is same as rejecting (n – r) objects.

\color{green} (4-) color(red)(text()^nC_a = text()^(n)C_b) => a =b  or a = n - b , i.e., color(blue)(n =a + b)

\color { maroon} ® \color{maroon} ul (" REMEMBER")

color(red)(text()^nC_r +text()^nC_(r-1)=text()^(n+1)C_r )

"Proof" We have : text()^nC_r+text()^nC_(r-1) = (n!)/(r! (n- r)!) + ( n!)/((r-1)! ( n-r+1)!)

 = (n!)/(r xx (r-1)! ( n-r)!) + (n!)/((r-1) ! ( n-r+1)(n-r)!)

= (n!)/((r-1)! (n-r)!) [1/r +1/(n-r+1)]

 = (n!)/((r-1) !(n-r)!) xx (n -r +1+r)/(r(n-r+1)) = ((n+1)!)/(r!(n+1-r)!) = text()^(n+1)C_r

Combination/selection/collection/committee/team refers to the situation where order of occurrence of the
event is not important.

Q 3048445303 If text()^nC_9 = text()^nC_8 find n Solution:

We have text()^nC_9 = text()^nC_8

(n!)/{9! (n-9)!} = (n!)/{(n-8)! 8!}

or 1/9 = 1/(n-8)

or n-8=9

or n = 17

Therefore text()^nC_(17) = text()^(17)C_(17) = 1
Q 3028545401 A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Solution:

Here, order does not matter. Therefore, we need to count combinations. There will be as many committees as there are combinations of 5 different persons

taken 3 at a time. Hence, the required number of ways = text()^5C_3 = (5!)/(3! 2!) = (4xx5)/2 = 10.

Now, 1 man can be selected from 2 men in 2C_1 ways and 2 women can be selected from 3 women in 3C_2 ways. Therefore, the required number of committees

 = text()^2C_1 xx text()^3C_2 = (2!)/(1! 1!) xx (3!)/(2! 1!) = 6
Q 3058645504 What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these
(i) four cards are of the same suit,
(ii) four cards belong to four different suits,
(iii) are face cards,
(iv) two are red cards and two are black cards,
(v) cards are of the same colour? Solution:

There will be as many ways of choosing 4 cards from 52 cards as there are combinations of 52 different things, taken 4 at a time. Therefore

The required number of ways  = text()^(52)C_4 = (52 !)/(4 ! 48!) = (49xx50xx51xx52)/(2xx3xx4)

 = 270725

(i) There are four suits: diamond, club, spade, heart and there are 13 cards of each suit. Therefore, there are text()^(13)C_4 ways of choosing 4 diamonds. Similarly, there are text()^(13)C_4 ways of choosing 4 clubs, text()^(13)C_4 ways of choosing 4 spades and text()^(13)C_4 ways of
choosing 4 hearts. Therefore

The required number of ways = text()^(13)C_4 +text()^(13)C_4+text()^(13)C_(14) +text()^(13)C_4

 = 4xx(13 !)/(4! 9!) = 2860

(ii) There are 13 cards in each suit.
Therefore, there are text()^(13)C_1 ways of choosing 1 card from 13 cards of diamond, text()^(13)C_1 ways of choosing 1 card from 13 cards of hearts, text()^(13)C_1 ways of choosing 1 card from 13 cards of clubs, text()^(13)C_1 ways of choosing 1 card from 13 cards of spades. Hence, by multiplication principle, the required number of ways

 = text()^(13)C_1 xx text()^(13)C_1 xx text()^(13)C_1 xx text()^(13)C_1 = 13^(4)

(iii) There are 12 face cards and 4 are to be slected out of these 12 cards. This can be done in text()^(12)C_4 ways. Therefore, the required number of ways = (12 !)/(4! 8!) = 495

(iv) There are 26 red cards and 26 black cards. Therefore, the required number of ways = text()^(26)C_2 × text()^(26)C_2

((26!)/(2! 24!))^2 = (325)^2 = 105625

(v) 4 red cards can be selected out of 26 red cards on text()^(26)C_4 ways. 4 black cards can be selected out of 26 black cards in text()^(26)C_4ways.

Therefore, the required number of ways  = text()^(26)C_4 +text()^(26)C_4

 = 2xx (26!)/(4! 22 !) = 29900 