Exams Joint Entrance Examination Mathematics Permutations and Combinations Question 23 Question 23 of 278 Hard MCQ
Let n C r − 1 = 28 , n C r = 56 {}^{n}{C}_{r-1}=28,{}^{n}{C}_{r}=56 n C r − 1 = 28 , n C r = 56 and n C r + 1 = 70 {}^{n}{C}_{r+1}=70 n C r + 1 = 70 . Let A ( 4 c o s t , 4 sin t ) , B ( 2 sin t , − 2 cos t ) A(4cost,4\sin t),B(2\sin t,-2\cos t) A ( 4 cos t , 4 sin t ) , B ( 2 sin t , − 2 cos t ) and C ( 3 r − n , r 2 − n − 1 ) C(3r-n,{r}^{2}-n-1) C ( 3 r − n , r 2 − n − 1 ) be the vertices of a triangle A B C ABC A B C , where t t t is a parameter. If ( 3 x − 1 ) 2 + ( 3 y ) 2 (3x-1{)}^{2}+(3y{)}^{2} ( 3 x − 1 ) 2 + ( 3 y ) 2 = α =\alpha = α , is the locus of the centroid of triangle ABC , then α \alpha α equals
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n C r − 1 = 28 , n C r = 56 n C r − 1 n C r = 28 56 n ! ( r − 1 ) ! ( n − r + 1 ) ! n ! r ! ( n − r ) ! = 1 2 r ( n − r + 1 ) = 1 2 3 r = n + 1 \begin{gathered} {}^{n}{C}_{r-1}=28,{}^{n}{C}_{r}=56 \\ \frac{{}^{n}{C}_{r-1}}{{}^{n}{C}_{r}}=\frac{28}{56} \\ \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}}=\frac{1}{2} \\ \frac{r}{(n-r+1)}=\frac{1}{2} \\ 3r=n+1\, \end{gathered} n C r − 1 = 28 , n C r = 56 n C r n C r − 1 = 56 28 r ! ( n − r )! n ! ( r − 1 )! ( n − r + 1 )! n ! = 2 1 ( n − r + 1 ) r = 2 1 3 r = n + 1
n C r n C r + 1 = 56 70 ( r + 1 ) ( n − r ) = 56 70 ⇒ 9 r = 4 n − 5 \begin{gathered} \frac{{}^{n}{C}_{r}}{{}^{n}{C}_{r+1}}=\frac{56}{70} \\ \frac{(r+1)}{(n-r)}=\frac{56}{70}\Rightarrow 9r=4n-5\, \end{gathered} n C r + 1 n C r = 70 56 ( n − r ) ( r + 1 ) = 70 56 ⇒ 9 r = 4 n − 5 Answer the question to unlock