It's time to solve Combinatorics

Probability Level 3

If C ( 2 n , 2 ) = a C ( n , 2 ) + n b \, C(2n,2) = a C(n,2) + n^b , where a a and b b are integers and C ( j , k ) = j ! k ! ( j − k ) ! C(j,k) = \dfrac{j!}{k!(j-k)!} is the binomial coefficient , find a b + a b a^b+ ab .

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The answer is 8.

Akshay Sharma by Akshay Sharma , 21, India

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1 solution

Chew-Seong Cheong
Apr 25, 2016

C ( 2 n , 2 ) = a C ( n , 2 ) + n b ⟹ ( 2 n ) ! 2 ! ( 2 n − 2 ) ! = a n ! 2 ! ( n − 2 ) ! + n b 2 n ( 2 n − 1 ) 2 = a n ( n − 1 ) 2 + n b 4 n 2 − 2 n = a n 2 − a n + 2 n b \begin{aligned} C (2n, 2) & = a C (n, 2) + n^b \\ \implies \frac{(2n)!}{2!(2n-2)!} & = \frac{a n!}{2!(n-2)!} + n^b \\ \frac{2n(2n-1)}{2} & = \frac{an(n-1)}{2} + n^b \\ 4n^2 - 2n & = an^2 -an + 2n^b \end{aligned}

Equating the coefficients, we have: { b = 1 ⟹ a = 4 b = 2 ⟹ a = 2 \begin{cases} b =1 & \implies a = 4 \\ b = 2 & \implies a = 2 \end{cases}

Therefore a b + a b = { 4 1 + 4 × 1 2 2 + 2 × 2 = 8 a^b + ab = \begin{cases} 4^1 + 4 \times 1 \\ 2^2 + 2 \times 2 \end{cases} = \boxed{8}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

a = b = 2 also works ( doesn't affect the answer )

A Former Brilliant Member - 5 years, 1 month ago

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Yes, you are right.

Chew-Seong Cheong - 5 years, 1 month ago

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