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00:00 - 00:59 | hello friend here we have a question that and boys and girls sit alternatively along a line in X ways and along a circle in Y ways such that x is equal to 10 why this is given to us than the question is asking that the number of ways in which and boys can sit around the table so that none of the has the same two neighbours that their neighbour should not be same right so hair number of ways to set and boys and girl alternatively in a line so number of ways will be number of ways to Said boys and girls boys and girls alternatively ulta |

01:00 - 01:59 | alternatively in a line in a line this will be equal to that is let's suppose the boy said first N in between them the girls will sit let's suppose boy sit here here here here and so on and in between girl will set right so this can be done in that boys will sit in and place so that is and then they can arrange in and factorial way and girls will sit in NC dinner and factorial weight now there here we start with boys hair girls can also start right so we will get to ke so we will multiply with to this is given to us X right now in the circle that is number of |

02:00 - 02:59 | ways number of ways to Said boys and girls alternatively ulta natively in a circle so when they sit in a circle that is circle right Sohail let's suppose boys it Hare Hare Hare Hare Hare Hare right so boys can be arranged in and -1 factorial with bright as we know the number of arrangement of anything in a circular will be and -1 factorial write an in between them girls can set the girls will sit in an places in a n factorial with right so this will be an factorial way no hair |

03:00 - 03:59 | weaving multiply with two because we don't know from where they start because when when the people sitting in a circle there is not a starting for leather repair the starting point Nora ending point so we can multiply with to know this is given to us why right now 10 act that is the given is X equal to 10 why this implies that to into n factorial into and factorial equal to N -1 factorial into n factorial and into 10 right and Fact and fact we will cancel out here this will cancel out with 5 now n factorial this and factorial we can write and into and -1 factorial here and -1 factorial |

04:00 - 04:59 | 25 now this and this will cancel out so we got the value of n is 5 so that moment we have 5 boys and 5 girls now 5 boys can arrange in a circle that they won't have a same neighbour to let suppose we have a circle right so this is a circle have 5 boy can set that is 12345 right now 5 boys can fit in a circular Infotech tutorial way right but info factorial when when there is arrangement in this side that support this is a boy right and they will have a same they will have a same neighbour that is suppose A and B they will have a neighbour A and B so when they were |

05:00 - 05:59 | in this direction when the going distraction in this direction again they have a same name but that is ab but their order is changed that is we will come here and we will come here right so we what we will do we will divide by 4 factorial by to write so we won't get the same case again so total arrangement will be total arrangement of boys total arrangement will be 5 - 1 factorial divided by 2 that is 4 factorial divided by 24 factorial value is equal to 24 / to this is comes as 12 so answer is 12 thank you |

**Why we need Complex Number ?**

**Algorithm to find integral exponents of iota and generalize in terms of 4n+1 ; 4n; 4n+2**

**Definition Of Complex Numbers**

**Equality of complex numbers**

**Addition of complex number and their properties**

**Subtraction of complex numbers**

**multiplication of two complex no. and their properties**

**Division of two complex number**

**Conjugate of a complex no and its properties. If `z, z_1, z_2` are complex no.; then :-
(i) `bar(barz)=z` (ii)`z+barz=2Re(z)`(iii)`z-barz=2i Im(z)` (iv)`z=barz hArr z` is purely real (v) `z+barz=0implies` z is purely imaginary (vi)`zbarz=[Re(z)]^2+[Im(z)]^2`**

**Properties of a complex no. If `z;z_1;z_2` are complex no.; then (vii)`bar(z_1+z_2)=barz_2+barz_1` (viii)`bar(z_1-z_2)=barz_1-barz_2` (ix)`bar(z_1z_2)=barz_1barz_2` (x) `(barz_1)/z_2=barz_1/barz_2` where `z_2!=0`**