Advertisements
Advertisements
प्रश्न
Compute: `(8!)/(6! - 4!)`
Advertisements
उत्तर
`(8!)/(6! - 4!) = (8 xx 7 xx 6 xx 5 xx 4 xx 3 xx 2 xx 1)/((6 xx 5 xx 4 xx 3 xx 2 xx 1) - (4 xx 3 xx 2 xx 1))`
= `40320/(720 - 24)`
= `40320/696`
= `1680/29`
∴ `(8!)/(6! - 4!) = 1680/29`
= 57.93
APPEARS IN
संबंधित प्रश्न
Evaluate: 8!
Evaluate: 10!
Evaluate: 10! – 6!
Compute: (3 × 2)!
Compute: 3! × 2!
Compute: `(9!)/(3! 6!)`
Compute: `(8!)/((6 - 4)!)`
Write in terms of factorial.
5 × 6 × 7 × 8 × 9 × 10
Evaluate : `("n"!)/("r"!("n" - "r")!)` for n = 8, r = 6
Evaluate `("n"!)/("r"!("n" - "r")!)` for n = 15, r = 10
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 (n + 3)! = 110 × (n + 1)!
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
Find n, if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 7)!)` = 1 : 6
Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24 : 1
Simplify `((2"n" + 2)!)/((2"n")!)`
Simplify `(("n" + 3)!)/(("n"^2 - 4)("n" + 1)!)`
Simplify `1/("n"!) - 1/(("n" - 1)!) - 1/(("n" - 2)!)`
Simplify n[n! + (n – 1)!] + n2(n – 1)! + (n + 1)!
Simplify `1/(("n" - 1)!) + (1 - "n")/(("n" + 1)!)`
Simplify `("n"^2 - 9)/(("n" + 3)!) + 6/(("n" + 2)!) - 1/(("n" + 1)!)`
Find the number of integers greater than 7,000 that can be formed using the digits 4, 6, 7, 8, and 9, without repetition: ______
If `((11 - "n")!)/((10 - "n")!) = 9,`then n = ______.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon of n sides. If Tn + 1 – Tn = 21, then n is equal to ______.
