यदि `((2n)!)/(3!(2n-3)!)` तथा `(n!)/(2!(n-2)!), 44:3` के अनुपात में हैं, तो n का मान ज्ञात कीजिए ।

Explanation

प्रश्नानुसार
`((((2n)!)/(3!(2n-3)!)))/(((n!)/(2!(n-2)!)) ) = 44/3`
`rArr ((2n)!2!(n-2)!)/(3!(2n-3)!n!) = 44/3`
`rArr (2n(2n-1)(2n-2)(2n-3)!)/(3!(2n-3)!) .(2!(n-2)!)/(n(n-1)(n-2)!)= 44/3`
`rArr(2n(2n-1)(2n-2))/(1.2.3) xx (1.2)/(n(n-1)) = 44/3`
`rArr (2n-1)(2n-2) = 22(n-1)`
`rArr 4n^(2) - 6n+ 2 = 22n - 22`
`rArr 4n^(2) -28+24 = 0`
`rArr n^(2 ) - 7n+ 6 =0 rArr(n-1) (n-6) = 0`
`rArr n = 1` या `n = 6`