Advertisements
Advertisements
प्रश्न
Show that: `(9!)/(3!6!) + (9!)/(4!5!) = (10!)/(4!6!)`
उत्तर
L.H.S. = `(9!)/(3!6!) + (9!)/(4!5!)`
= `(9!)/(3!xx6 xx 5!) + (9!)/(4 xx 3! xx 5!)`
= `(9!)/(5!3!) [1/6 + 1/4]`
= `(9!)/(5! xx 3!)[(4 + 6)/(6xx4)]`
= `(9!xx10)/(6xx5!xx4xx3!)`
= `(10!)/(6!4!)`
= `(10!)/(4!6!)`
= R.H.S.
APPEARS IN
संबंधित प्रश्न
Evaluate: 8! – 6!
Compute: `(9!)/(3! 6!)`
Compute: `(6! - 4!)/(4!)`
Write in terms of factorial:
5 × 6 × 7 × 8 × 9 × 10
Write in terms of factorial:
6 × 7 × 8 × 9
Write in terms of factorial:
5 × 10 × 15 × 20 × 25
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 8, r = 6
Find n, if `"n"/(8!) = 3/(6!) + 1/(4!)`
Find n, if `"n"/(6!) = 4/(8!) + 3/(6!)`
Find n, if `1/("n"!) = 1/(4!) - 4/(5!)`
Find n, if (n + 1)! = 42 × (n – 1)!
Find n, if: `((17 - "n")!)/((14 - "n")!)` = 5!
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24:1
Find the value of: `(5(26!) + (27!))/(4(27!) - 8(26!)`
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
A student passes an examination if he/she secures a minimum in each of the 7 subjects. Find the number of ways a student can fail.
How many quadratic equations can be formed using numbers from 0, 2, 4, 5 as coefficient if a coefficient can be repeated in an equation.
