USERSPPB

THERE ARE 5 CASES

1ST CASE-1-1-1-1-0-0-0-0,TOTAL POSSIBLE =4*3*2*1*(8!/4!/4!(FOR BOX))=24*70=1680.

2ND CASE-2-1-1-0-0-0-0-0,TOTAL POSSIBLE =4C2*2*(8!/5!/2!)=2016.

3RD CASE-2-2-0-0-0-0-0-0,TOTAL POSSIBLE =4C2*2C2*8!/6!/2!=336.

4TH CASE-3-1-0-0-0-0-0-0,TOTAL POSSIBLE =4C3*1*8!/6!=224.

5TH CASE-4-0-0-0-0-0-0-0,TOTAL POSSIBLE=4C4*8!/7!=1*8=8.

TOTAL WAYS=1680+2016+336+224+8=4264.

USERSPPB

WE CAN WRITE THE EQUATION LIKE \(\begin{pmatrix} 1&&2&&1\\ 2&&-1&&2\\ 3&&1&&-1 \end{pmatrix} \begin{pmatrix} x\\ y\\ z \end{pmatrix} = \begin{pmatrix} 5\\ 15\\ 8 \end{pmatrix}\)

SOLVING THE EQUATIONS WE GOT X=4, Y=-1 , Z = 3

so, our answer is \(\begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix}\).

USER SPPB

PLEASE SEND THE DIAGRAMS TO SOLVE THE PROBLEM 

PLEASE SEND THE DIAGRAM TO SOLVE THE QUESTION.

answer is 12\(\sqrt2\).

explain :

DRAW A LINE X-B-C AND CX PERPENDICULAR TO AX .

\( \angle\)ABC=135 .THEN \( \angle\)ABX=45=\(\angle\)BAX.(\(\angle\)AXB=90)

45:45:90=1:1:\(\sqrt2\)

AB=6 ,SO AX=BX=6/\(\sqrt2\)=3\(\sqrt2\).

BC=8.

AREA  OF TRIANGLE ABC=(1/2)*AX*BC=(1/2)*3\(\sqrt2\)*8=12\(\sqrt2\).

This is simple geometry. The triangles ABC,BXC,AXB are all similar. You are given AX=12,XC=4=12. So it is easy to find the other lengths. Eg XC/BC=BC/AC, so BC=8 and hence BX=4sqrt3.