Permutations and Combinations (Part - 1) - Math, Class 11 Video Lecture

okay the next topic is the next is
combinatorics which is the part of
permutation and combination that is P
and C let us start there first problem
that is given to us as DP p of p and c
permutation and combination or you may
say it as combi networks occasion one of
this DPP is find the number of ways
and which
12 students
may be equally
divided into three groups we have to
find the number of phase in which 12
students may be equally divided into
three groups to solve the vocation we
have assuming that
the group's are
distinguishable
distinguishable the answer may be given
as
12 c 4 into 8 c 4 that means 12
factorial upon cube of 4 factorial here
first of all from among 12 we have make
a group of four students then they are
remaining eight students then we have
taken four strands from these errands
and the remaining will make a third
group next question question number two
is given to us as see mr. Mohan invites
a party
of 15 guests to our dinner
and places
ten of them
at a round table
and another five
at another round table
then find the total number of
with in which he can arrange
all the guests
solution of the given equation may be
given as he can choose
10 people
to be seated
at the first table
in 15 c 10 where's the chosen 10 can be
arranged
round the first table
in 9 factorial ways as we know that
there are n minus 1 factorial ways to
sit and number of people around a table
now the remaining five
can be arranged
around the second table
in poor vectorial ways so the required
number is
15c 10 into 9 factorial into 4 factorial
that is 16 factorial upon 10 in 25 there
it is our answer moving to our next
question question number 3 that is a
lift with
seven people
stops at ten floors from zero to
seven
people go out at each floor
how many ways
are there for the lift to be empty
here the ways only differ
by the number of people
leaving at
each block
let us have the solution of the given
equation the solution may be given as we
have to require to find the number of
integer solution of x1 plus x2 plus x3
up to X 7 or X 10 is equal to 7 where x
is greater than equal to 0 so the answer
will be given as
7 plus 10 minus 1 c 7 that is 16 c 7 31
and so we will do our next question
question number 4 that is a
parallelogram
is cut by two sets of M lines parallel
to its sides then find the number of
parallelograms
then bomb
to solve a vocation we have as we can
see that there are M plus two parallel
lines
in one direction
and M plus two parallel lines
in another direction
including
the sides
a parallelogram
is formed
by two lines
choosing from each group
so the number of l'a grams
is n plus 2 C 2 into n plus 2 C 2 moving
to our next question cashier number five
that is given to us as in how many ways
in how many ways
can the candidates
a1 a2 a3 up to a 10 B then
so that
hey one is always above a two to solve
the equation let us have the solution as
let n be the required number
of ways on each way
we can interchange we can interchange
the ranks of
a-one a-two without altering
the ranks
first the total number of
ways of distributing
the Rings is 2n but 10 ranks can be
distributed can be distributed among ten
candidates
in 10 factorial ways
it follows that n is equal to 10
factorial upon to that is are unsure
next is question number six that is in
how many ways
can an odd number of objects
be selected
wrong and objects
for solution of vocation we have since
since the number of ways in which
our object
can be selected is it an see are so the
required
number is and c 1 plus NC three plus NC
5 plus dash dash dash that may be given
as 2 days 2 power n minus 1 that is up
moving to our next question question
number 7 next question number 7 is let n
is equal to 9 + 99 + 9 99 + dash dash
dash up to 99 999 where the last
some and
consists of 1999 visits of nine then
tell how many times
will the desert
one appear and and this all of occasion
we have n is given as nine that means
10-1 99 that means hundred minus 1999
that means thousand minus one first dash
dash dash up to 10 raised to power one
triple 9-1 that they are two series 10
plus 10 square plus 10 cubed plus dash
dash dash up to 10 raised to power one
triple nine minus one plus one plus 14
one triple nine times so we have the
first is the GP poo some may be given as
1 1 1 1 1 1 0 minus 1999 thatfull that
will give us 1 11 11 10 9 11 1 so the
number of one is the required number is
number of one is 2000-2009 so down there
is 1998