In how many ways 6 boys and 6 girls can be arranged along a row  so that (i) all the girls come together

(ii) no two girls come together (iii) no two of the same sex come together

(i) 6boys and 6 girls are arranged along a row So that all girls come to gether:
consider 6 girls as one unit . 6 boys and unit of girls $B_1,B_2,B_3,B_4,B_5,B_6,G\text{. }$canbe arranged in 7! and 6 girts can be arrarged in 6!
no.of arrangements =7! x 6!

(ii) no two girls come togethe 6 boys can be arranged in 6! ways 
$-B_1-B_2-B_3-B_4-B_5-B_6-$ there are 7 places in ${ }^7{P}_6$ ways ${.}^7{P}_6=7!$
number of arrangements =$=7!\times 6!=7\times 6!\times 6!$

(iii) no two of the same sex come togeter 6 boys , 6 girls can be arranged alternatevely in 2 ways and 6boys can be arranged among them in 6! , 6 girls can be rearranged in 6! ways 
no.of arrangements $=2\times 6!\times 6!$