"Day 50 of My Learning Journey: Setting Sail into Data Excellence! Today's Focus: Maths for Data Analysis ( Per & Comb- 5)

PERMUTATION AND COMBINATION - 5

Permutation = arrangement of object
Combination = selection of object

Different Object & Alike Object

Different object -
Eg :- a,b,c,d.

Alike Object -
Eg :- a,a,a,a.

Permutation = arrangement of object
Types :-
Permutation / Arrangement of Different object
Eg :- Permutation / arrangement of ‘n’ different objects is given a factorial of ‘n’.
Permutation / Arrangement of Alike object
Eg :- permutation of ‘n’ different objects is given a factorial of ‘n’.

Combination = selection of object
Types :-
Combination / Selection of Different object
Eg :- permutation of ‘n’ different objects is given a factorial of ‘n’.
Combination / Selection of Alike object
Eg :- permutation of ‘n’ different objects is given a factorial of ‘n’.

NUMERICAL

Q :- 4 BOYS & 4 GIRLS are to be arranged in a straight line. In how many ways this can be done.
Answer the following and fulfil the condition.

Without any restriction

It means 8 objects = 8!

If all boys are together

Box method : - 
Boys can be arranged 4! ( internal arrangement inside box )

Girl + boy box = 5!

Answer = 4! * 5!

If all boys are together & all girls are also together.

Box method : - 
Boys can be arranged 4! ( internal arrangement inside box )
Girls can be arranged 4! ( internal arrangement inside box )

Girl box + boy box = 2!

Answer = 4! * 2! * 2!

If boys are separated ( no two boys are together )

GAP method : - 
Girls can be arranged 4! 
BOYS are arranged in Gap = ( total Gap = 5 ) 

Answer = 4! * 5*4*3*2

If no two boys are together , no two girls are together.
GAP method : -
Girls can be arranged 4!
BOYS are arranged in 4!
Gap must be maintain = 2

Answer = 4! * 2 * 4!

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