Let $${z_k}$$ = $$\cos \left( {{{2k\pi } \over {10}}} \right) + i\,\,\sin \left( {{{2k\pi } \over {10}}} \right);\,k = 1,2....,9$$

List-I

P. For each $${z_k}$$ = there exits as $${z_j}$$ such that $${z_k}$$.$${z_j}$$ = 1
Q. There exists a $$k \in \left\{ {1,2,....,9} \right\}$$ such that $${z_1}.z = {z_k}$$ has no solution z in the set of complex numbers
R. $${{\left| {1 - {z_1}} \right|\,\left| {1 - {z_2}} \right|\,....\left| {1 - {z_9}} \right|} \over {10}}$$ equals
S. $$1 - \sum\limits_{k = 1}^9 {\cos \left( {{{2k\pi } \over {10}}} \right)} $$ equals

List-II

1. True
2. False
3. 1
4. 2

The quadratic equation $$p(x)$$ $$ = 0$$ with real coefficients has purely imaginary roots. Then the equation $$p(p(x))=0$$ has

Coefficient of $${x^{11}}$$ in the expansion of $${\left( {1 + {x^2}} \right)^4}{\left( {1 + {x^3}} \right)^7}{\left( {1 + {x^4}} \right)^{12}}$$ is

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can be done is