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Probability Level 2

If mC2= n then find the value of (m+1)C4?

(nC3)/2 (nC2)*3 (nC2) (nC2)/3

Gulshan Sharma by Gulshan Sharma , 23, India

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1 solution

Vivek Pattanayak
Dec 17, 2017

mC2 = m ( m − 1 ) 2 \frac {m(m-1)}{2} =n ---(1)

And (m+1)C4= ( m + 1 ) ( m ) ( m − 1 ) ( m − 2 ) 4 ! \frac {(m+1)(m)(m-1)(m-2)}{4!} ;

Also, in equation 1 subtracting 1 from both sides,

( m ( m − 1 ) 2 \frac{m(m-1)}{2} ) -1=n-1

=> m 2 − m − 2 2 \frac {m^2 -m -2} {2} =n-1;

=> ( m + 1 ) ( m − 2 ) 2 = n − 1 \frac {(m+1)(m-2)}{2}=n-1 ;-----(2)

Multiplying equation 1 and 2;

( m + 1 ) ( m ) ( m − 1 ) ( m − 2 ) 2 ⋅ 2 = n ( n − 1 ) \frac {(m+1)(m)(m-1)(m-2)}{2\cdot2}=n(n-1)

Dividing both sides by 2x3;

( m + 1 ) ( m ) ( m − 1 ) ( m − 2 ) 4 ( 2 ⋅ 3 ) = 1 3 ( n ( n − 1 ) / 2 ) \frac {(m+1)(m)(m-1)(m-2)}{4(2 \cdot 3)}= \frac{1}{3} (n(n-1)/2)

=> ( m + 1 ) ( m ) ( m − 1 ) ( m − 2 ) 4 ! = 1 3 ( n C 2 ) \frac{(m+1)(m)(m-1)(m-2)}{4!} = \frac {1}{3}(nC2)

=> (m+1)C4= 1 3 ⋅ \frac {1}{3} \cdot nC2

Hence proved

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