DevSquare Quantitative Aptitude Permutation and Combination Questions

Say M came first. The remaining 4 positions can be filled in 4! = 24 ways.

Now M came in second. N can finish the race in 3rd, 4th or 5th position. So total ways are 3 x 3! = 18.

M came in third. N can finish the race in 2 positions. 2 x 3! = 12.

M came in second. N can finish in only one way. 1 x 3! = 6

Total ways are 24 + 18 + 12 + 6 = 60.

10C1*5C1*8C1*2C1*21C2=168000
IN EVERY FIELD ONE MEMBER IS SELECTED AND IN REMAINING 21 WE SELECT 2 MEMBER.

No. of zeros means to check what is the total no. of factors of 2 and 5 when no. is divided by both. consider lowest no. of factors
i.e. 99/2=49, 49/2=24, 24/2=12, 12/2=6, 6/2=3, 3/2=1.

Therefore total no. of factors of 2=49+24+12+6+3+1=95
now 99/5=19, 19/5=3.

therefore total no. of factors =19+3=22
now take the lowest one....so 22 no. of zero in 99!

five girls can be seated in circle in (5-1)! ways=4!=24
remaining boys
can be placed in 5! ways=120
total ways =120*24=2880

since no two women can come together,therefore women can be
selected in 3c1 ways ...and men will be chosen in 12c6 so to
complete the 7 group members...
=12_{C_{6}}\times3_{C_{1}}

and only selection of men also possible =12_{C_{7}}

so final ans :

12_{C_{6}}\times3_{C_{1}}+12_{C_{7}}

34 ways

One boy and one girl can be selected as (one boy & three girls) or
(two boys & two girls) or (three boys & one girl)=
(4_{C_{1}}\times3_{C_{3}})+(4_{C_{2}}\times3_{C_{2}})+(4_{C_{3}}\times3_{C_{1}})=4+18+12=34

d)none of these
4 men can take 4 position which can be done in 4!
and women
can be sit in 3! ways so 4!*3!

or

Given that the total number of people = 7
First we select 4 positions for the men to occupy.
These positions could be chosen in 7C4 ways = 7!/4!(7-4)! =
(1 x 2 x ... x 7)/(1 x 2 x 3 x 4)(1 x 2 x 3)
= (5 x 6 x 7)/(1 x 2 x 3)
= 35 ways.

This leaves 3 positions for the women, and the women can
be rearranged amongst themselves in 3! ways = 6 ways.
The men, of course, can be in only one order amongst themselves,
so the overall number of ways they could line up is given by
35 x 6 = 210 ways.
Hence, the answer is 210

as Cube has 3 dimensions i.e L,B,H

now 1 cut on any side divide cube into 2 parts

similarly 4 cuts on any side ( say on L) will divide it in 5 Parts
Hence 4 cuts on B will make ( 5*5 =25 parts ) and 4 cuts on
H ( 5*5*5 = 125 )
Therefore
4+4+4 =12 cuts required

total item : 6
we can select : 4

thus -> 6C4 which = 6C2
:: 6*5/ 2*1 = 15

ans 12
if one face have 1,3,2, then the crossponding face will be occupied
by respective pair i.e (6,4,5).
so there r 3 place and we have 3 place soit can be arranged in 3! way
i.e 6 way and every pair will be suffled in 2 way so ans will be6*2=12.