Quantitative Aptitude – Permutation and Combination
One of the most important chapter of many competitive examination is Permutation and Combination. Questions are often comes from this chapter. Exams like govt, banking and other private sector, must have a question from this chapter.
Permutation and Combination is not a very hard to learn. You just need to concentrate on basics of this chapter. To understand this chapter well you need to learn Factorial first. Factorial is very much needed to solve Permutation and Combination problems. So, now we will discuss the basic ideas and formulas of this chapter.
Let’s starts the discussion.
What is Permutation ?
Definition: When the order of a group of objects is important, we can call it Permutation.
Example:
Suppose, you have a 4 digit numbered pin lock in your Mobile phone and the unlock pattern is 1234. Therefore here the exact pin sequence is very much important to unlock the phone. If we enter 4321 or any other number sequence, we will not unlock the phone. But the phone will only unlock if we enter 1234, the exact sequence. So, this is the perfect example of Permutation.
Formula of Permutation
The formula of all possible permutations of “r” objects from a set of “n” objects.
nPr = n (n – 1) (n – 2)…(n – r + 1)
= n! / (n – r)!
Therefore, if we want to find the permutation of 8P3. We will do,
So, here, n = 8 and r = 3. And also (n – r + 1) = (8 – 3 + 1) = 6 equivalent to (8 – 2).
∴ 8P3 = 8 (8 – 1) (8 – 2)
= 8 X 7 X 6
= 336
What is Combination ?
Definition: When the order of a group of objects is not important, we can call it Combination.
Example:
Suppose, you need to pay Rs.35/- to a grocery shop. You have three notes. A 20 rupee note, a 10 rupee note and a 5 rupee note. Since it doesn’t really matter how you arrange those note. Whether you arrange it 20-10-5 or you arrange it 5-10-20 or 10-5-20. So, when the order is not the important thing in arranging something, we can call it Combination.
Formula of Combination
The formula of all possible combinations of “r” objects from a set of “n” objects.
nCr = n! / r! (n – r)!
= n (n – 1) (n – 2)…(n – r + 1) / r!
Therefore, if we want to find the combination of 8C3. We will do,
So, here, n = 8 and r = 3. And also (n – r + 1) = (8 – 3 + 1) = 6 equivalent to (8 – 2).
∴ 8C3 = 8 (8 – 1) (8 – 2) / 3!
= (8 X 7 X 6) / (3 X 2 X 1)
=336 / 6
= 56
So, this is the basic ideas and formula of permutation and combination. Hope you understand this chapter. Now if you have any questions in your mind then you can ask us here in the comment section of www.AptitudeTricks.com. Our team member will discuss your question here.