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Permutations and Combinations question

2013 · Shift 0 · Q40
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Permutations and Combinations question

2013 · Shift 0 · Q40

JEE MainMathematicsPermutations and CombinationsMCQ+4 / −1
Let Tn{T_n}Tn​ be the number of all possible triangles formed by joining vertices of an n-sided regular polygon. If Tn+1−Tn{T_{n + 1}} - {T_n}Tn+1​−Tn​ = 10, then the value of n is :
  1. A
    7
  2. B
    5
  3. C
    10
  4. D
    8
View written solutionFree

Correct answer: B

  1. Number of triangles from an nnn-gon

Any triangle is formed by choosing any 333 vertices from the nnn vertices of the polygon. Thus,

Tn=(n3)=n(n−1)(n−2)6.T_n = {n \choose 3} = \frac{n(n-1)(n-2)}{6}.Tn​=(3n​)=6n(n−1)(n−2)​.

Similarly,

Tn+1=(n+13)=(n+1)n(n−1)6.T_{n+1} = {n+1 \choose 3} = \frac{(n+1)n(n-1)}{6}.Tn+1​=(3n+1​)=6(n+1)n(n−1)​.
  1. Compute the difference

Given:

Tn+1−Tn=10.T_{n+1} - T_n = 10.Tn+1​−Tn​=10.

Now,

(n+13)−(n3)=(n2).{n+1 \choose 3} - {n \choose 3} = {n \choose 2}.(3n+1​)−(3n​)=(2n​).

So,

(n2)=10.{n \choose 2} = 10.(2n​)=10.

That is,

n(n−1)2=10.\frac{n(n-1)}{2} = 10.2n(n−1)​=10.
  1. Solve for nnn
n(n−1)=20n(n-1) = 20n(n−1)=20 n2−n−20=0n^2 - n - 20 = 0n2−n−20=0 (n−5)(n+4)=0(n-5)(n+4)=0(n−5)(n+4)=0

So,

n=5orn=−4.n=5 \quad \text{or} \quad n=-4.n=5orn=−4.

Since number of sides must be positive,

n=5.n=5.n=5.
  1. Check with options

Option BBB is 555, so the correct answer is:

5\boxed{5}5​
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