Question

How do you simplify \[\dfrac{{21!}}{{17!4!}}\] ?

Answer 1

Hint: Here this problem is from factorials generally related to permutations and combinations. Here we are given the factorials in the form of a fraction. We will write the 21! In the form of expansion such that after 17! The value would be the same. So don’t expand after 17!. Also keep 4! As it is or do write its value to cancel it with the product of numbers remaining above. So let’s start!

Complete step by step solution:
Given that
 \[\dfrac{{21!}}{{17!4!}}\]
We know that 21! Is the product of numbers in decreasing order upto 1 starting from 21, but we will keep numbers after 17 as it is so as to cancel it with the 17! In the denominator.
 \[21! = 21 \times 20 \times 19 \times 18 \times 17!\]
Now putting this value in the numerator of the fraction above.
 \[ = \dfrac{{21 \times 20 \times 19 \times 18 \times 17!}}{{17!4!}}\]
Cancelling 17! From numerator and denominator,
 \[ = \dfrac{{21 \times 20 \times 19 \times 18}}{{4!}}\]
Now let’s expand 4!,
 \[ = \dfrac{{21 \times 20 \times 19 \times 18}}{{4 \times 3 \times 2 \times 1}}\]
Now we need to divide and simplify the numbers, divide 21 by 3, divide 20 by 4
 \[ = \dfrac{{7 \times 5 \times 19 \times 18}}{2}\]
Now divide 18 by 2 we get,
 \[ = 7 \times 5 \times 19 \times 9\]
On multiplying we get,
 \[ = 5985\]
So, the correct answer is “5985”.

Note: Here the thing that one should note is don’t take the value of 21! directly. That is too big a number to calculate. Instead the denominator will help in cancelling the big product. Whereas the denominator has 19! So we will write the numerator upto 19! And then only products of 21 and 20 in the numerator will be there.