यदि `.^(n)C_(r-1)=36,^(n)C_(r) = 84"और" ^(n)C_(r+1)=126,"तो "^(r)C_(2)` का मान ज्ञात कीजिए ।

Explanation

`(.^(n)C_(r-1))/(.^(n)C_(r)) = 36/84`
` rArr (n!)/((r-1)!(n-r+1)!) *(r!(n-r)!)/(n!) = 3/7`
` rArr (r(r-1)!(n-r)!)/((r-1)!(n-r+1)*(n-r)!) = 3/7`
` rArr r/(n-r+1) = 3/7`
` rArr 3n - 3r + 3 = 7r`
` rArr n = (10r-3)/3` ...(1)
और ` (.^(n)C_(r))/(.^(n)C_(r+1)) = 84/126`
` rArr (n!)/(r!(n-r)!) ((r+1)!(n-r-1)!)/(n!) = 2/3`
` rArr ((r+1)* r!(n-r-1)!)/(r !(n-r)(n-r-1)!) = 2/3`
` rArr (r+1)/(n-r) = 2/3`
` rArr 2n - 2r = 3r + 3`
` rArr n = (5r+3)/2`
...(2)
समीकरण (1) और (2) से
` (5r+3)/2 = (10r-3)/3`
` rArr 20r - 6 = 15r + 9`
` rArr r =3`
अब ` .^(r)C_(2) = .^(3)C_(2)`
` = (3!)/(2! *1!) = 3`.