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प्रश्न
Show that: `(9!)/(3!6!) + (9!)/(4!5!) = (10!)/(4!6!)`
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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.
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संबंधित प्रश्न
If numbers are formed using digits 2, 3, 4, 5, 6 without repetition, how many of them will exceed 400?
Evaluate: 6!
Evaluate: (8 – 6)!
Compute: `(12!)/(6!)`
Compute: `(12/6)!`
Compute: `(9!)/(3! 6!)`
Compute: `(8!)/(6! - 4!)`
Compute: `(8!)/((6 - 4)!)`
Evaluate: `("n"!)/("r"!("n" - "r"!)` For n = 12, r = 12
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"!)/(3!("n" - 5)!) : ("n"!)/(5!("n" - 7)!)` = 10:3
Find n, if: `((17 - "n")!)/((14 - "n")!)` = 5!
Find n, if: `((15 - "n")!)/((13 - "n")!)` = 12
Show that
`("n"!)/("r"!("n" - "r")!) + ("n"!)/(("r" - 1)!("n" - "r" + 1)!) = (("n" + 1)!)/("r"!("n" - "r" + 1)!`
Find the value of: `(8! + 5(4!))/(4! - 12)`
Show that: `((2"n")!)/("n"!)` = 2n(2n – 1)(2n – 3)....5.3.1
A hall has 12 lamps and every lamp can be switched on independently. Find the number of ways of illuminating the hall.
