When using the pumping lemma with length to prove that the language $L=\{b_m{b}^n,b0\}$ is nonregular, the following approach is taken.

Assume $L$ is regular. Then there exists an FA with $k$ states which accepts $L.$

We choose a word $w=b{a}^k{b}^k=xyz,$ which is a word in L$.$ Some options for chosing $xyz$ exist.

$x=,y=b{a}^p,z=b^{k-p},$ for some $x=b,y=a^p,z=a^{k-p}{b}^{k}x=b{a}^k,y=a,z=b^{k}x=b{a}^p,y=a^{k-p}b,z=b^{k-p-q}x=,y=b{a}^q,z=a^{k-p-a}{b}^{k}p,q>0,p+q$

Which one $of$ the following would $be$ the correct set $of$ options $to$ choose?

Select one:

$a.1$ and $2$

$b.1,3,$ and $5$

$c.1,2,$ and $4p,q>0,p+q$

$x=,y=b{a}^q,z=a^{k-p-a}{b}^k,$ for some $p,q>0,p+q$

Which one $of$ the following would $be$ the correct set $of$ options $to$ choose?

Select one:

$a.1$ and $2$

$b.1,3,$ and $5$

$c.1,2,$ and $4p>0,p$

$x=b{a}^k,y=a,z=b^k$

$x=b{a}^p,y=a^{k-p}b,z=b^{k-p-q},$ for some $p,q>0,p+q$

$x=,y=b{a}^q,z=a^{k-p-a}{b}^k,$ for some $p,q>0,p+q$

Which one $of$ the following would $be$ the correct set $of$ options $to$ choose?

Select one:

$a.1$ and $2$

$b.1,3,$ and $5$

$c.1,2,$ and $4p>0,p$

$x=b,y=a^p,z=a^{k-p}{b}^k,$ for some $p>0,p$

$x=b{a}^k,y=a,z=b^k$

$x=b{a}^p,y=a^{k-p}b,z=b^{k-p-q},$ for some $p,q>0,p+q$

$x=,y=b{a}^q,z=a^{k-p-a}{b}^k,$ for some $p,q>0,p+q$

Which one $of$ the following would $be$ the correct set $of$ options $to$ choose?

Select one:

$a.1$ and $2$

$b.1,3,$ and $5$

$c.1,2,$ and $4$