Q. 184.2( 5 Votes )
Prove that : 4nC2n : 2nCn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.
Answer :
Given that we need to prove:
4nC2n : 2nCn = [1 . 3 . 5 …. (4n – 1)] : [1.3.5….. (2n – 1)]2.
Consider L.H.S:
We know that 
And also n! = n(n – 1)(n – 2)…………2.1
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
= R.H.S
∴ L.H.S = R.H.S, thus proved.
Rate this question :
Permutations & Combinations | Analyze your learningFREE Class
Challenging Quiz on P&C | Test Yourself55 mins
Understand Permutations like never before60 mins
Check Your progress Part 2| Interactive Quiz: Permutation & Combination54 mins
Interactive Quiz on Combinations50 mins
Lecture on Combinations49 mins
Interactive Quiz on Division and distribution of objects17 mins
Division and distribution of objects58 mins
Interactive Quiz on Combinations-0253 mins
Circular permutations61 mins
Prove that the product of 2n consecutive negative integers is divisible by (2n)!
RD Sharma - MathematicsThere are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.[Hint: Required number = 210 – 1].
Mathematics - ExemplarIf n+1Cr+1 : nCr = 11 : 6 and nCr : n–1Cr–1 = 6 : 3, find n and r.
RS Aggarwal - MathematicsIf nCr–1 = 36, nCr = 84 and nCr+1 = 126, find r.
RS Aggarwal - MathematicsFill in the Blanks
If nPr = 840, nCr = 35, then r = ______.
Mathematics - ExemplarThe number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
Mathematics - ExemplarTotal number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
Mathematics - Exemplar