Below are detailed answers for each of the questions asked in test, the explanations given below would further strengthen your understanding of Permutation & Combinations topic.
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Q1. How
many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which
are divisible by 5 and none of the digits is repeated?
Answer:
Since
each desired number is divisible by 5, so we must have 5 at the unit place. So,
there is 1 way of doing it.
The tens
place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So,
there are 5 ways of filling the tens place.
The
hundreds place can now be filled by any of the remaining 4 digits. So, there
are 4 ways of filling it.
Required
number of numbers = (1 x 5 x 4) = 20.
Q2.In how
many ways a committee, consisting of 5 men and 6 women can be formed from 8 men
and 10 women?
Answer:
Required
number of ways = (8C5 x 10C6)
= (8C3 x
10C4)
= 8 x 7 x 6 x 10 x 9 x 8 x 7
3 x 2 x 1 4 x 3 x 2 x 1
= 11760.
Q3.In how
many ways can 4 girls and 5 boys be arranged in a row so that all the four
girls are together??
Answer:
Let 4
girls be one unit and now there are 6 units in all.
They can
be arranged in 6! ways.
In each
of these arrangements 4 girls can be arranged in 4! ways.
=>
Total number of arrangements in which girls are always together
=6!*4!=720*24=
17280
Q4. 12
points lie on a circle. How many cyclic quadrilaterals can be drawn by using
these points?
For any set of 4 points we get a cyclic quadrilateral. Number of ways of
choosing 4 points out of 12 points is 12C4=495
.
Therefore, we can draw 495 quadrilaterals.
Q5. The
Indian Cricket team consists of 16 players. It includes 2 wicket keepers and 5
bowlers. In how many ways can a cricket eleven be selected if we have to select
1 wicket keeper and at least 4 bowlers?
We are to choose 11 players including 1 wicket keeper and 4 bowlers
or, 1
wicket keeper and 5 bowlers.
Number of
ways of selecting 1 wicket keeper, 4 bowlers and 6 other players
=2C1*5C4*9C6=840
Number of
ways of selecting 1 wicket keeper, 5 bowlers and 5 other players
=2C1*5C5*9C5=252
=>
Total number of ways of selecting the team:
=840+252=
1092
Q6. How
many factors of (2^4)×(5^3)×(7^4) are odd numbers?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be an odd number, a should be 0.
b can take values 0,1, 2,3, and c can take values 0, 1, 2,3, 4.
Total number of odd factors =4×5= 20
Q7. How
many factors of (2^5)×(3^6)×(5^2) are perfect squares?
Any factor of this number should be of the form 2a×3b×5c.
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2.
Total number of perfect squares =3×4×2= 24
Q8. In
how many ways can eight directors, the vice-chairman and chairman of a firm be
seated at a round table, if the chairman has to sit between the vice-chairman
and the director?
In a circular arrangement problem, we always fix one position and the look at
ways of arranging the rest with respect to the fixed position.
Case 1 -
seating of chairman - In this question, we fix the position of the Chairman.
Thus the Chairman can be seated in 1 way only.
Case 2 -
seating of vice-chairman - The vice-chairman can be to the left or right of the
chariman. Thus the vice chairman can be seated in 2 ways.
Case 3 -
seating of the 8 directors - The rest of the 8 directors can be seated in 8
positions in 8! ways. This is because the first director can be seated in andy
of 8 positions in 8 ways. And then the second director can be seated in any of
the remaning 7 positions in 7 ways and so on, thus giving the total number of
ways of seating the 8 director as 8 * 7 * 6* 5 * 4 * 3 * 2 * 1, i.e. 8! ways.
Thus the
desired answer is when case 1 and 2 and 3 happen together, which is 1 * 2 * 8!
= 2 * 8! ways.
Q9. In
how many ways can 15 people be seated around two round tables with seating
capacities of 7 and 8 people?
‘n' objects can be arranged around a circle in (n - 1)!.
If
arranging these 'n' objects clockwise or counter clockwise means one and the
same, then the number arrangements will be half that number.
i.e.,
number of arrangements = (n-1)!/2.
You can
choose the 7 people to sit in the first table in 15C7 ways.
After
selecting 7 people for the table that can seat 7 people, they can be seated in
(7-1)! = 6!.
The
remaining 8 people can be made to sit around the second circular table in
(8-1)! = 7! Ways.
Hence,
total number of ways 15C8 * 6! * 7!
Q10.
Suppose you can travel from a place A to a place B by 3 buses, from place B to
place C by 4 buses, from place C to place D by 2 buses and from place D to
place E by 3 buses. In how many ways can you travel from A to E?
The bus from A to B can be selected in 3 ways.
The bus
from B to C can be selected in 4 ways.
The bus
from C to D can be selected in 2 ways.
The bus
from D to E can be selected in 3 ways.
So, by
the General Counting Principle, one can travel from A to E in 3*4*2*3= 72 ways
Q11. How
many words can be formed by using 4 letters at a time of word
"SURPRISE" (words may be meaningful or meaningless according to
dictionary)
(1) No. of words having all letters different = 6*5*4*3 = 360
(2) No.
of words having any one letter repeated
=(2*1)*(5*4*3*2*1/3*2*1*2*1)*(4*3*2*1/2*1)
= 240
(3) No.
of words having two letter repeated = (1)*(4*3*2*1/2*1*2*1) = 6
Total
words = 360+240+6 = 606
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