Advertisements
Advertisements
प्रश्न
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
Advertisements
उत्तर
L.H.S. = `((2"n")!)/("n"!)`
`= ((2"n")(2"n"-1)(2"n"-2)(2"n"-3)(2"n"-4) ...6xx5xx4xx3xx 2xx1)/("n"!)`
`=((2"n")(2"n" - 1)[2("n" - 1)](2"n" - 3)[2("n" - 2)]...(2xx3)xx5xx(2xx2)xx3xx(2xx1)xx1)/("n"!)`
`=(2^"n"["n"("n"-1)("n"-2)....3.2.1][(2"n"-1)(2"n"-3)...5.3.1])/("n"!)`
= `(2^"n"("n"!)(2"n"-1)(2"n"-3)...5.3.1)/"n!"`
= 2n(2n −1)(2n −3)...5.3.1
= R.H.S.
APPEARS IN
संबंधित प्रश्न
A teacher wants to select the class monitor in a class of 30 boys and 20 girls. In how many ways can he select a student if the monitor can be a boy or a girl?
How many three-digit numbers can be formed using the digits 2, 3, 4, 5, 6 if digits can be repeated?
Evaluate: 8!
Evaluate: (8 – 6)!
Compute: (3 × 2)!
Compute: 3! × 2!
Compute: `(8!)/((6 - 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
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 12, r = 12
Find n if: `("n"!)/(3!("n" - 3)!) : ("n"!)/(5!("n" - 5)!)` = 5:3
Find n if: `("n"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
A question paper has 6 questions. How many ways does a student have if he wants to solve at least one question?
