MATHEMATICS DBT 2

Prob 1

Ans : B ?????????????

bhaiya one of the unsolved is wat rkrish
2 has edited .. pls help wid dat

bhaiya one of the unsolved is wat rkrish
2 has edited .. pls help wid dat

someone please help HELP!!!!!! the SUMS

well its x_n+1 not x_n+k

Is this the question ????

Prob 3

x_{n+1} = \frac{1+x^{n}}{1-x^{n}} for all n \epsilon N and x_{k}=x_{1}=35

Find k.

See at ur own risk !!

Put n=4

Total no. of ways = 4! = 24

a) 6(16-27+2) < 0 XXX

b) 12(9-8+1) = 24 √

c) 24(3-1) - 24(4-4) = 48 X

I solved it by some shortcut which I don't know is useful and correct always or not.

Playpower, in prob 1: does each urn contain n balls of each color or is total number of balls is n?

WHERE BELOW?

please post the solution of sum no. 1

is the ans B??

for the first one?

PROBLEM 1
n balls are picked one by one from 4 urns each containing n identical balls of green , red , white and blue colours . The number of ways , when no 2 consecutive pickings are from the same urn and balls of all the colours are picked , are
a> 6(4^{n-2}-3^{n-1}+ 2^{n-3})
b>12(3^{n-2}-2^{n-1}+ 1)
c>24(3^{n-3}-2^{n-4})-24(n-4)
d> none of these

Can u pls latexify the other questions.They're not clearly visible.

YUP THNKS i got i [1]
@ rkrish
pls try other sums

I hope you got this :
N = 2⁴ . 5⁴ . ......

After this ,

you have to get

(p₁+1)(p₂+1)(p₃+1)....(p_n+1) = 105

105 = 3.5.7 = (2+1).(4+1).(6+1) √
= 15.7 = (14+1).(6+1) √
= 5.21 = (4+1).(20+1) √
= 3.35 = (2+1).(34+1) X (this is not possible as we know that N = 2⁴ . 5⁴ . ......... )

So, at the most 3 prime factors are possible (2,5 & a)
Hence, the cases mentioned above.

am i rite Manish bhaiyya ??????

Prob 4

N = a₁^p₁ . a₂^p₂ . a₃^p₃ . .... . a_n^p_n

(p₁+1)(p₂+1)(p₃+1)....(p_n+1) = 105 (which is odd)

So, p₁,p₂,p₃, ... ,p_n should all be even.

\Rightarrow N should be a perfect square.

N = 2³ . 5³ . ...........

So, definitely N is also divisible by 10000 (since powers of prime factors are even).

N = 2⁴ . 5⁴ . ...........

Now, 105 = 3 . 5 . 7

So, power of 2 & 5 cannot be 4 simultaneously.

So,
either N = 2⁴ . 5⁶ . a²
or N = 2⁶ . 5⁴ . a²
or N = 2⁴ . 5²⁰
or N = 2²⁰ . 5⁴
or N = 2⁶ . 5¹⁴
or N = 2¹⁴ . 5⁶

Therefore,
Total no. of odd factors can be = 12 , 8 , 20 , 4 , 14 or 6

Hence options A,B,C,D are correct.