**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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cant understand Q11 solution!! can anyone pls explain

ReplyDeleteHi PalaneSami,

DeleteBelow is the detailed answer:

List & count of alphabets which are part of word SURPRISE:

E=1, I=1, P=1, R=2, S=2 & U=1

which is 6 unique letters (EIPRSU - Lets represent any letter of this unique set as X) and 2 repeated letters(R & S - lets represent any letter of this repeated set as Y).

We are asked to form different 4 letter words using words in surprise. Looking at the composition of letter of the word surprise this could be done as follows:

1.XXXX - Using all unique letters i.e. all 4 letters from EIPRSUNo. of words having all letters different = 6*5*4*3 = 360

2.XXXY - Using 3 unique and 1 repeated letter i.e.2 letters from EIPRU and 2 letters of S OR

2 letters from EIPSU and 2 letters of R

Hence 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.XXYY - Using all letters from repeated letters set i.e. all 4 from RRSS setHence 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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