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GI Quant: Pre – Counting – Combinations theory + examples

combination is a way of choosing elements from a set in which order does not matter.

Consider the following example: Lisa has 12 different ornaments and she wants to give 5 ornaments to her mom as a birthday gift (the order of the gifts does not matter). How many ways can she do this?

We can think of Lisa giving her mom a first ornament, a second ornament, a third ornament, etc. This can be done in 12!/7! ways. However, Lisa’s mom is receiving all five ornaments at once, so the order Lisa decides on the ornaments does not matter. There are 5! re-orderings of the chosen ornaments, implying the total number of ways for Lisa to give her mom an unordered set of 5 ornaments is 12!/(7!5!)

Q. How many ways are there to arrange 3 chocolate chip cookies and 10 raspberry cheesecake cookies into a row of 13 cookies?

We can consider the situation as having 13 spots and filling them with 3 chocolate chip cookies and 10 raspberry cheesecake cookies. Then we just choose 3 spots for the chocolate chip cookies and let the other 10 spots have raspberry cheesecake cookies. 

The number of ways to do this job is :

13C3 = (13 x 12 x 11)/ (3 x 2 x 1) = 286

Note: Even if we chose 10 spots for the raspberry cheesecake cookies instead, the answer will be the same.

13C10 = 13C3 = 13!/(10!x3!)

Q. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

The number of ways of selecting 3 consonants out of 7 and 2 vowels out of 4 is

7C3 x 4C2 = 210.

Therefore, the number of groups each containing 3 consonants and 2 vowels is 210. Since each group contains 5 letters, which can be arranged amongst themselves in 5! = 120 ways, the required number of words is 210 x 120 = 25200

Q. How many ways are there to select 3 males and 2 females out of 7 males and 5 females?

The number of ways to select 3 males out of 7 is

7C3 = (7 x 6 x 5)/ (3 x 2 x 1) = 35

Similarly, the number of ways to select 2 females out of 5 is

5C2 = (5 x 4)/ (2 x 1) = 10

Hence, we can select 3 males AND 2 females in: 35 x 10=350 ways. 

Q. . There are 9 children. How many ways are there to group these 9 children into three groups of 2, 3, or 4 children?

It matters whether a child joins a group of 2 or a group or 3 or 4.

So, we could break down the problem into three components:

i. # of ways to select 2 individuals from 9
ii. # of ways to select 3 from remaining 7
iii. # of ways to select 4 from remaining 4

And then multiple these three results to get the total number of ways.

The number of ways to choose 2 children out of 9 is

9C2 = ( 9 x 8)/ (2 x 1) = 36

The number of ways to choose 3 children out of 9 – 2=7 is

7C3 = ( 7 x 6 x 5 )/ (3 x 2 x 1) = 35

Finally, the number of ways to choose 4 children out of 7-3=4 is

4C4 = 1

Hence, the answer is 36 x 35 x 1=1260 ways

Q. There are 9 distinct chairs. How many ways are there to group these chairs into 3 groups of 3?

The number of ways to choose 3 chairs out of 9 is

9C3 = (9 x 8 x7)/ (3 x 2 x 1)

 The number of ways to choose 3 chairs out of 9- 3 = 6 is

6C3 = ( 6 x 5 x 4)/ (3 x 2 x1)

Finally, the number of ways to choose 3 chairs out of 6-3= 3 is

3C3 = 1

Now, since each of these three groups has an equal number of three chairs and the order of the three groups does not matter, our answer is

(84 x 20 x 1) / 3! = 280

Q. At a party everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?

Answer: 12

Let’s start with 1 individual, and consider how many handshakes would be there. This may help us understand the path to take.

If 1 person attended the party, number of handshakes: 0
If 2 people attended the party, number of handshakes: 1
If 3 people attended the party, number of handshakes: 3 (First guy with second and third, second guy with third. That’s it.)
If 4 people attended the party, number of handshakes:

First person with: second, third and fourth: 3 handshakes
Since the second has already shaken hands with the first, we’ll only consider cases not counted yet.
Second person with: third and fourth: 2 handshakes
Third person with: fourth: 1 handshake

So, if 4 people attended the party, number of handshakes: 3 + 2 + 1 = 6

We’re seeing a trend.

Let’s say there are n people at the party. The first person shakes hands n-1 times, since he does not shake hands with himself. The next person then shakes hands with n-2 people (since he has already shaken hands with the first person), and so on until the last person has already shaken everyone’s hand by the time it’s his turn.

The total number of shakes for a party with n people is then given by 1+2+3+…+(n-1).

1+2+3+…+(n-1) = (n-1)(n)/2 (Sum of n terms formula. Here we have ‘n-1’ terms.)

Now, we are given that the total number of handshakes is 66.

i.e., (n-1)(n)/2 = 66

n^2 – n – 132 = 0
n^2 -12n + 11n – 132 = 0
n(n – 12) + 11(n – 12) = 0
(n-12)(n+11) = 0
n = 12 or -11.

Number of people can’t be negative. Therefore, n = 12.

Q. How many ordered non-negative integer solutions (a, b, c, d) are there to the equation a + b + c + d = 10?

We can think of this problem the same way as the ‘distributing 5 pens among 3 individuals’ problem.

First think of: in how many ways can 10 identical pens be distributed among 4 individuals A, B, C and D?

Now, once the distribution is done, let’s count the number of pens each individual has. Those numbers are basically the values of a, b, c and d. A can get any number of pens from 0-10. The total number of pens will remain 10. So, no matter how the pens are distributed, the total takes care of itself.

e.g, we could do: (10,0,0,0), (9,1,0,0), (9, 0, 1, 0), (9, 0, 0, 1), (8, 2, 0, 0) … and so on.

We could notice trends to answer, in how many ways can A get n pens.

In how many ways can A get 10 pens? 1
In how many ways can A get 9 pens? 3 (the last pen can go to B, C or D)
In how many ways can A get 8 pens? The remaining 2 pens can be distributed in 6 ways: (2,0,0), (1,1,0), (1,0,1), (0,2,0), (0,1,1), (0,0,2)
In how many ways can A get 7 pens? The remaining 3 pens can be distributed as follows: (3,0,0), (2,1,0), (2,0,1), (1,2,0), (1,1,1), (1,0,2), (0,3,0), (0,2,1), (0,1,2), (0,0,3): 10 ways

The difference is increasing by 2, 3, 4 and so on.

So, total number of ways will be: 1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 + 66 = 286.

Alternately,

This question can be answered using a technique known as stars and bars.

Say there are 10 stars that need to be distributed among 4 people.

One way to think about it is to put 3 bars (‘|’) somewhere in the middle to make three partitions and 4 portions.

e.g. * * | * * * | * * * * | *

This way A gets 2, B 3, C 4 and D 1.

So, the idea is to arrange 10 *’s and 3 |’s.

Basically, we need to arrange 13 characters such that 10 are identical, and the other 3 are identical.

This can be done in 13!/(10! x 3!) ways.

This gives us the answer: 286 such sequences.

Q.There are 5 shirts all of different colors, 4 pairs of pants all of different colors, and 2 pairs of shoes with different colors. In how many ways can Amy and Bunny be dressed up with a shirt, a pair of pants, and a pair of shoes each?

We choose 2 shirts out of 5 for both Amy and Bunny to wear, so

5C2 x 2! = 20

We choose 2 pairs of pants out of 4 for them to wear, so

4C2 x 2! = 12

We choose 2 pairs of shoes out of 2 for them to wear, so

2C2 x 2! = 2

Therefore, by the rule of product, the answer is 20×12×2=480 ways

Q. Find the number of rectangles in a 10×12 chessboard.

In an a x b board, there are a+1 horizontal lines and b+1 vertical lines. Since a rectangle has 2 horizontal and 2 vertical lines then the number of rectangles would be:

11C2 x 13C2 = 4290

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