Quiz 1

we can choose any 3 places for white marbles in 9c3 and remaining places are left for black marbles; we do not have to arrange them because they all are identical

For a number to be divisible by 8, its last three digits
must divisible by 8.

So, last three digits can be from the following group
(024), (048), (064), (208), (240), (248), (264), (280),
(408), (480), (608), (624), (640), (648), (680), (824), (840), (864),

There are 18 such groups.

Consider these 12 groups in which zero is there
(024), (048), (064), (208), (240), (280), (408), (480),
(608), (640), (680), (840)

With each of these 12 groups, the first digit can be any of
the remaining 2 digits and the second digit can be the
remaining one digit
Therefore, 12×2 = 24 such numbers

Consider the 6 groups in which zero does not come
(248), (264), (624), (648), (824), (864),

With each of these 6 groups, the first digit can be the one
remaining digit(because zero cannot be the first digit)
The second digit can be zero
Therefore, 6×1 = 6 such numbers

Therefore, total count of 5 digit numbers = 24 + 6 = 30
So, Ans is 30

The answer would be 56.
Lets say we have removed the three said houses. Now we
are left with 7 houses.
If we arrange these 7 houses we have 8 places to put
the other 3 houses at.

_ H1 _ H2 _ H3 _ H4 _ H5 _ H6 _ H7 _

so 8C3 = 56

total number arrangement without any restriction is 2^5=32
nymber of arrangement in which blue balls are adjacent =bb,bb,bb,bb,bbb,bbb,bbb,bbbb,bbbb,bbbbb=10

hence required number= 32-10=22

if you fix the 2nd and 3rd position which can be done in
10 ways and 1st position can take value from 1-9 therefore
total ways comes out to b 9*10=90
plus
if you fix the 1st nd 2 position which can be done in 9 ways and 3rd position can take 10 values i.e from 0-9, that makes it 9*10=90

total=90+90=180
now minus the common part like 111,222 to 999
that is
total of 9 numbers..
therefore finally 180-9=171 which is the answer

Case 1: when there is 2 25pc coin - 1 ways

Case 2: when there is 1 25pc coin and
(1) 2 10pc coin - nos of 5pc coin will range from 1 to 0 i.e 2 combinations
(2) 1 10pc coin - nos of 5pc coin will range from 3 to 0 i.e 4 combinations
(3) 0 10pc coin - nos of 5pc coin will range from 5 to 0 i.e 6 combinations

Case 3: when there is no 25 pc coins and
(1) 5 10pc coin - 1 ways
(2) 4 10pc coin - nos of 5pc coin will range from 2 to 0 i.e 3 combinations
(3) 3 10pc coin - nos of 5pc coin will range from 4 to 0 i.e 5 combinations
(4) 2 10pc coin - nos of 5pc coin will range from 6 to 0 i.e 7 combinations
(5) 1 10pc coin - nos of 5pc coin will range from 8 to 0 i.e 9 combinations
(6) 0 10pc coin - nos of 5pc coin will range from 10 to 0 i.e 11 combinations

So a total 49 different combinations.

he either can can have 2 chocolates and 1 icecream or 1 chocolate and 2 icecream.
so the solution would be.
10C2*12C1 + 10C1*12C2 = 1200

if total seats N = 5 in a circular arrangement ( _ _ M _ _ M ) where
first ( _ ) is connected with last Man(M), then min no of persons
required= 2.
similarly If N = 9, Man required = 3 i.e,( _ _ M _ _ M _ _ M )

If N= 12, Man required = 4 i.e, ( _ _ M _ _ M _ _ M _ _ M)
have u got any relation??
the no of man required is = Total no of seats in circle / 3
So, A/Q total seat = 60. Man required = 60/3=20 Ans.

The number of ways of distributing $N$ identical objects among $R$ group such that each group can have any number of objects is given by:

$C(N+R-1,R-1)=C(10+6-1,6-1)$

$=C(15,5)$

$=15C5$

7C5*3C2=63