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Explain in Detail :Simplifying Factorial Expressions
A factorial expression is an algebraic expression that is composed of the product of all positive integers from 1 to a given number. Factorial expressions can be simplified by cancelling out common factors. For example, the factorial expression 5! can be simplified to 5*4*3*2*1.
Simplifying Negative Integer Expressions
To simplify a negative integer expression, divide the expression by -1.
Problem solving is perhaps the best way to see mathematics in action. So here we are to take you through the problems that will significantly help you learn how to simplify factorial expressions.
In order to simplify factorial expressions using variables that’re there in the numerator and denominator— we would require to produce common factors in both locations such that they can be canceled. Because, this is of course our objective.
The key is to establish a comparison between the factorials as well identify which amongst them is higher in value. Let’s say we seek to compare the factorials ({n + 4}!(n+4)! And {n + 2}!(n+ 2)!
It is pretty obvious to observe that {n + 4}! > {n + 2}! Is true for all values of ‘n’ inasmuch as the factorial is interpreted, that is to say, the items in the interior of the parenthesis is a whole number ≥ zero (0).
Thus, this suggests that we can stretch {n + 4}! Until the expression {n+2}! Takes shape into the sequence.
Like this below:-
(N+4)! = (N+3). (N+3) (N+2)!
Simplifying Negative Integer
Let’s take for example {n – 5}! And {n – 2}! In such a situation, we are subtracting the variable by some number. Keep in mind that here the bigger expression is the one with smaller subtrahend, or the value that’s subtracted from the minuend. Hence, this suggests that {n – 2}! > {n – 5}!
This further concludes that the expression {n – 2}! Will have {n – 5}! In its expanded form
(N -2)! = (N-2). (N-3) . (N-4) . (N-5)!
Solved Example
Problem 1
How many different ways can 7 students in a relay race come 1st, 2nd and 3rd?
Solution 1
Seeing that it is a long-drawn list, thus, if 7 students are named T, U, V, W, X, Y, Z
Then the list involves:
TUV, TUW, TUX, TUY, TUZ, TVU, TVW, TVX, TVY… etc
Using the formula – 7! (7−3)! = 7! 4!
Next, we will write the multiplications out in full:
7* 6 * 5 * 4 * 3 * 14 * 3 * 2 * 1 = 7 * 6 * 5
That was neat. The multiplication of {4 × 3 × 2 × } “invalidated”, giving up only 7 × 6 × 5. And:
7 * 6 * 5 = 210
Thus, as an answer, we get 210 different ways by which 7 students secure positions at 1st, 2nd and 3rd.
And you’re done!
Problem 2
How many ways can we arrange 4 alphabets from the start (without repeating)?
Solution 2
In order to determine the equals, you need to simply multiply the positive integers altogether that are less than or equal to (≤) 4. Now, likewise
For 1 Alphabet “a” there is solely 1 way: 1×1 = (a)
For 2 alphabet “ab” there are 2 ways: 1×2= (ab, ba)
For 3 Alphabet “abc” there are 6 ways: 1×2×3 = (abc, acb, bac, bca, cab, cba)
For 4 Alphabets “abcd” there are 24 ways: 1×2×3×4= (abcd, acbd, adbc, adcb, abdc, acdb, bacd, bcad, bcda, bdac, bdca, badc, cabd, cbad, cbda, cdab,cdba, cadb, dabc, dacb, dbac, dbca, dcab, dcba)
