GRE QUANTITATIVE: PERMUTATIONS, COMBINATIONS AND PROBABILITY

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What are permutations?
What are combinations?
How do we use them in selection and arrangements?
Learn all this and much more in this presentation.

Number of ways of arranging r objects out of n objects = 𝑛 𝑃𝑟
=
𝑛!
𝑛−𝑟 !
5 𝑃2
= 5 4 = 20
Number of ways of selecting r out of n objects = 𝑛 𝐶 𝑟
=
𝑛!
𝑟! 𝑛−𝑟 !
5 𝐶2
=
5(4)
1(2)
= 10

Question 1
6 points lie in a circle. How many quadrilaterals can be formed by joining these
points.

We need to choose 4 out of 6 points since a quadrilateral has 4 vertices.
Total number of quadrilaterals =6 𝐶4
= 6 𝑐2
=
6×5
1×2
= 15

Question 2
Quantity A
All the jacks are removed from a pack of cards and then a card is drawn. Find the
probability of drawing a red king
Quantity B
2 unbiased dice are thrown. Find the probability of obtaining 6 as a sum or product

Quantity A
There are 4 jacks which are removed. There are 48 cards left . We need to draw 1
red king from 4 kings.
Probability =
4 𝐶1
48 𝐶1
=
4
48
=
1
12

Quantity B
We need 6 as a sum or product
Favourable outcomes are = {(1,5), (1,6),(2,4),(2,3), (3,3),(3,2),(4,2),(5,1),(6,1)}
𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
9
36
=
1
4
A B
1
12
1
4
Cross multiplying, we get
A B
4 12
B is greater

Question 3
Compare the 2 quantities
Quantity A
In how many ways can an examiner select a set of 6 true/false conditions
Quantity B
In how many ways can an examiner enter a set of 5 objective questions,
each question having 3 choices

Option A
There are 3 options to the true/false question
ie true, false or not answered
Since there are 6 questions, total number of ways =36
Option B
There are 5 questions, each question having 3 choices
Total number of ways = 35
Option A is greater

Question 4
An urn contains 5 blue, 5 red, 5 green and 5 yellow balls. Find the probability
that if 2 balls are drawn at random, both are blue.

Bl R G Y
5 5 5 5
𝑃robability =
5 𝐶2
20 𝐶2
=
5
65

Question 5
In how many ways can 5 children be seated in a row so that 2 children
never sit together

Treat the 2 children as one.
There are now 5 -2 +1 =4 children
These can be arranged in 4! =24 ways
Number of arrangements of n objects = n!
The 2 children within themselves can be seated in 2 ways
Total number of ways = (24)(2)=48

Question 6
How many 5 digit numbers can be formed using 0,2,3,4 and 5 such that repetition
is allowed and the number is divisible by 2 or 5

The last digit can be 0,2 or 5.
The last digit can be filled in 3 ways --- --- ----- ----- -----
The first digit cannot be 0
The first digit can be filled in 4 ways
Since there is no restriction, the remaining 3 digits can be filled in 53 𝑤𝑎𝑦𝑠(5 × 5 × 5)
Answer = 125(3)(4)=1500 ways

Question 7
Team A has 5 athletes chosen from 7 people
Team B has 6 athletes chosen from 8 people
Column A Column B
Number of unique teams in A Number of unique teams in B

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑞𝑢𝑒 𝑡𝑒𝑎𝑚𝑠 𝑖𝑛 𝐴 = 7 𝐶5
=
7 × 6
1 × 2
= 21
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑢𝑛𝑖𝑞𝑢𝑒 𝑡𝑒𝑎𝑚𝑠 𝑖𝑛 𝐵 = 8 𝑐6
=
8 × 7
1 × 2
= 28
Column B is greater

Question 8
If the odds in favour of an event are 4:5, find the probability that it will occur.

𝑜𝑑𝑑𝑠 𝑖𝑛 𝑓𝑎𝑣𝑜𝑢𝑟 𝑜𝑓 𝑎𝑛 𝑒𝑣𝑒𝑛𝑡 ℎ𝑎𝑣𝑖𝑛𝑔 probability p =
𝑝
1−𝑝
𝑜𝑑𝑑𝑠 𝑎𝑔𝑎𝑖𝑛𝑠𝑡 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛𝑡 =
1 − 𝑝
𝑝
𝑝
1 − 𝑝
=
4
5
𝑠𝑜𝑙𝑣𝑖𝑛𝑔 , 𝑤𝑒 𝑔𝑒𝑡 𝑝 =
4
9

Question 8
The probability that A solves the problem is
1
9
𝑎𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡
B solves it is
5
18.
. 𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑜𝑛𝑙𝑦 1 𝑝𝑒𝑟𝑠𝑜𝑛 𝑠𝑜𝑙𝑣𝑒𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑙𝑒𝑚

Required answer = P(A𝐵) + 𝑃 𝐴𝐵 =
1
9
13
18
+
8
9
5
18
=
53
18

Question 9
Given 8 flags of different colours, how many different signals can be made using 3 flags
at a time one below the other

The first flag can be arranged in 8 ways.
Since they are of different colours, the 2nd flag can be arranged in7 ways.
The 3rd flag can be arranged in 6 ways.
Total number of ways = 8 × 7 × 6 = 336

Question 10
An urn contains balls numbered 6 to 20. Find the probability that the ball drawn is a multiple
of 4 or 5
Multiples of 4 : 8, 12, 16,20
Multiples of 5 : 10,15,20
A- multiple of 4 , 𝑃 𝐴 =
4
15
B: multiple of 5, P(B)=
3
15
𝑃 𝐴 ∩ 𝐵 =
1
15
Multiple of 4 and 5 : 20

P(A∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
=
4
15
+
3
15
−
1
15
=
6
15
=
2
5

Question 11
An urn contains 5 red, 6 blue, and 1 green ball. Another urn contains 6 red,
8 blue and 2 green balls. One ball is drawn at random from each urn. Find the
probability that both are green.

R BLUE G
5 6 1
R B G
6 8 2
P(𝐺1 𝐺2) = 𝑃 𝐺1 𝑃(𝐺2)=
1
12
2
16
=
1
96

What did we learn?
How to apply permutations
and
Combinations to solve
problems

GRE QUANTITATIVE: PERMUTATIONS, COMBINATIONS AND PROBABILITY

GRE QUANTITATIVE: PERMUTATIONS, COMBINATIONS AND PROBABILITY

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GRE QUANTITATIVE: PERMUTATIONS, COMBINATIONS AND PROBABILITY