मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी विज्ञान (सामान्य) इयत्ता ११ वी

Simplify 1n!-3(n+1)!-n2-4(n+2)! - Mathematics and Statistics

Advertisements
Advertisements

प्रश्न

Simplify `1/("n"!) - 3/(("n" + 1)!) - ("n"^2 - 4)/(("n" + 2)!)`

बेरीज
Advertisements

उत्तर

`1/("n"!) - 3/(("n" + 1)!) - ("n"^2 - 4)/(("n" + 2)!)`

= `1/("n"!) - 3/(("n" + 1)"n"!) - (("n" - 2)("n" + 2))/(("n" + 2)("n" + 1)"n"!)`

= `1/("n"!)[1-3/("n" + 1)-("n" - 2)/("n" + 1)]`

= `1/("n"!)[("n" + 1 -3 - "n" + 2)/("n" + 1)]`

= `1/("n"!)xx0`

= 0

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 3: Permutations and Combination - Exercise 3.2 [पृष्ठ ५०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 3 Permutations and Combination
Exercise 3.2 | Q 10. (vii) | पृष्ठ ५०

संबंधित प्रश्‍न

Evaluate: 8!


Evaluate: 10! – 6!


Evaluate: (10 – 6)!


Compute: `(12/6)!`


Compute: (3 × 2)!


Compute: `(9!)/(3!  6!)`


Compute: `(8!)/((6 - 4)!)`


Write in terms of factorial.

5 × 6 × 7 × 8 × 9 × 10


Write in terms of factorial.

3 × 6 × 9 × 12 × 15


Evaluate : `("n"!)/("r"!("n" - "r")!)` for n = 15, r = 8


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: `((17 - "n")!)/((14 - "n")!)` = 5!


Find n, if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5 : 3


Find n, if: `((2"n")!)/(7!(2"n" - 7)!) : ("n"!)/(4!("n" - 4)!)` = 24 : 1


Show that `("n"!)/("r"!("n" - "r")!) + ("n"!)/(("r" - 1)!("n" - "r" + 1)!) = (("n" + 1)!)/("r"!("n" - "r" + 1)!)`


Show that `((2"n")!)/("n"!)` = 2n (2n – 1)(2n – 3) ... 5.3.1


Simplify `((2"n" + 2)!)/((2"n")!)`


Simplify n[n! + (n – 1)!] + n2(n – 1)! + (n + 1)!


Simplify `("n" + 2)/("n"!) - (3"n" + 1)/(("n" + 1)!)`


Simplify `1/(("n" - 1)!) + (1 - "n")/(("n" + 1)!)`


Simplify `("n"^2 - 9)/(("n" + 3)!) + 6/(("n" + 2)!) - 1/(("n" + 1)!)`


Select the correct answer from the given alternatives.

In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?


Eight white chairs and four black chairs are randomly placed in a row. The probability that no two black chairs are placed adjacently equals.


3. 9. 15. 21 ...... upto 50 factors is equal to ______.


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 ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×