We are all familiar with multiplication. The factorial notation is a symbol that we use to represent a multiplication operation. But it is more than just a symbol. In the space below we will see what the factorial notation is and how we can use it to make our calculations easier. Let us begin with the introduction of the factorial and then we will see some solved examples of the same.
The factorial notation comes in handy when you are arranging objects. Consider the following scenario that we shall use to use to define and introduce this notation. For example, you have ten balls. Each ball has a number marked on it. You also have ten slots that you have to fill with the balls. How many different ways can you fill these slots in?
The first slot can be filled in 10 ways because you have 10 different balls to fill it with. You can fill the second slot in 9 ways. Since one of the slots already has a ball in it. Similarly, we can fill the next slot in 8 ways and so on. What is the total number of ways we can arrange these 10 balls in ten slots? This will be got from the fundamental principle of counting. The total number of ways is 10×9×8×7×6×5×4×3×2×1.
For all such arrangements, we will see a similar pattern of multiplication. For example, for any number ‘n’, we can make n×(n-1)×(n-2)×(n-3)×(n-4)×(n-5)×…×3×2×1. This is where we use the factorial notation. We define the factorial of a positive integer as the product of the integer with all the numbers lesser than it all the way up to 1.
We define the factorial of a number as the product of consecutive descending natural numbers and represent it by !. For example, the factorial of 4 or 4! = 4×3×2×1. Similarly the factorial of 7 or 7! = 7×6×5×4×3×2×1. Similarly, we can find the factorials of all the positive integers. In the factorial notation, we define the factorial of 0 to be = 1. So 0! = 1. By convention, 0! = 1. Also 1! = 1. Then 2! = 2 ! 1 = 2 and 3! = 3 ! 2 ! 1 = 6. Likewise, 4! = 4 ! 3 ! 2 ! 1 = 24 and 5! = 5 ! 4 ! 3 ! 2 ! 1 = 120.
Browse more Topics under Permutation And Combination
- Number of Permutations
- Number of Combination
- Permutation and Combination Practice Questions
Solved Examples For You
Example 1: Aman has 12 balls that have different numbers on it. He makes all the possible arrangements for the 12 different balls. Shoaib also has 6 balls that he arranges in all the possible orders. What is the ratio of the arrangements that Aman makes to the number of arrangements that Shoaib makes?
Answer: We know that the number of arrangements that we can make for any ‘n’ number of objects is given by n factorial or n!. Since Aman is arranging 10 objects, he will be able to do it in 10! ways. Similarly, Shoaib has 6 different objects that he will arrange in 6! ways. The ratio will simply be = 10!/6!
We can write this as (10×9×8×7×6×5×4×3×2×1)/6!
In other words, we can say that the ratio is = (10×9×8×7×6!)/6! = (10×9×8×7) /1 = 5040:1
Example 2: Evaluate the following: (i) 14!/8! (ii) 12!/(3!)(5!)
Answer: (i) The calculations with factorials can be difficult. We should try to reduce the numerator or the denominator such that a factorial term cancels itself. For example, in the first example we can write:
(14×13×12×11×10×9×8!)/8! = 14×13×12×11×10×9 = 2162160.
(ii) We shall use the same method to simplify the second part. Here we have, 12!/(3!)(5!) which can be written as:
(12×11×10×9×8×7×6×5!)/(3×2×1)(5!) = 665280
Examples of Arrangements
Example 3: A compact disc has 10 songs. The random play feature will play all of these songs in an unknown permutation (i.e., in order, without repeats).
(a) How many possible permutations are there of these 10 songs?
(b) If you select only your 8 favourite tracks and then use the random play, then how many possible arrangements will there be of these 8 songs?
(c) If you only have time to hear 4 songs on random play, then how many possibilities are there for a playing of 4 different songs from the entire CD?
Solution. (a) Since all 10 songs are to be arranged in order without repeats (i.e., permuted), there are 10! = 3,628,800 possibilities.
(b) Now only 8 songs are to be permuted, so there are 8! = 40,320 possibilities.
(c) Now 4 songs chosen from a set of 10 are to be listed in order without repeats. So now there are 10×9×8×7 = 5,040 possibilities. Note that the value in (c) also is given by (10×9×8×7×6×5×4×3×2×1)/(6×5×4×3×2×1) = 10!/6!
The 10! in the numerator comes from the total number of 10 songs. The 6! in the denominator comes from the number of unused songs in the list.
Example 4: In a psychological word association test, a computer will randomly pick a letter from the alphabet (A – Z) without repeating letters. The subject will have to say the first word coming to mind that starts with that letter. If the test goes on for 16 letters, then how many possibilities are there for the list of letters?
Answer: Because letters are not being repeated, we compute the number of choices by 26×25×24×. . .× until we have multiplied 16 terms together. It would be much easier to use 26!/10! ≈ 1.111363×1020 (which is a lot of possibilities). The 26! comes from the total set of 26 letters, and 10! comes from the number of 10 unused letters in the list.
Q 1: How many ways can you arrange the alphabet of the English language, if you were to form all the words that have three alphabets in them?
A) 15 B) 156 C) 1560 D) 15600
Ans: D) 15600