If `n! = 3(n-1)!` then `n` is
`(a) 2`
`(b) 3`
`(c) -3`
`(d) 1/2`

If a=3^(-3)-3^3\ a n d\ b=3^3-3^(-3),\ t h e n a/b-b/a (a) 0 (b) 1 (c)-1 (d) 2

If ^(n+1)C_3=2.^nC_2, then n= (A) 3 (B) 4 (C) 5 (D) 6

The coefficients of x^n in the expansion of (1+x)6(2n) and (1+x0^(2n-1) are in the rtio of (A) 1:2 (B) 1:3 (C) 3:1 (D) 2:1

Find the value of ( 1 - 1/3 ) ( 1 - 1/4 ) ( 1 - 1/5 ) ---------- ( 1 - 1/n ) is ( a ) 2/n ( b ) 4/n ( c ) 3/n ( d ) 5/n

Value of   iota^n + iota^( n+1 ) + iota^( n+2 ) + iota^( n+3 ) is ( a ) 1 ( b ) 2 ( c ) 4 ( d ) 0

Value of   iota^n + iota^( n+1 ) + iota^( n+2 ) + iota^( n+3 ) is ( a ) 1 ( b ) 2 ( c ) 4 ( d ) 0

The sum of the series 1/(1!(n-1)!)+1/(3!(n-3)!)+1/(5!(n-5)!)+…..+1/((n-1)!1!) is = (A) 1/(n!2^n) (B) 2^n/n! (C) 2^(n-1)/n! (D) 1/(n!2^(n-1)

For all n inN,3*5^(2n+1)+2^(3n+1) is divisible by (A) 19 (B) 17 (C) 23 (D) 25

Using the s,p,d notations, describe the orbital with the following quantum numbers : (a) n = 1, l = 0 " " (b) n = 3, l = 2 (c ) n = 3, l = 1 " " (d) n = 2, l = 1 (e ) n = 4, l = 3 " " (f) n = 4, l = 2 .

If 2^(n+4)-2^(n+2)=3, then n is equal to a.0 b.2c.-2d.-1