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CONCEPT TEST: Permutation and Combination

1. Find the total number of distinct vehicle numbers that can be formed using two

letters followed by two numbers. Letters need to be distinct.

A) 60000

B) 65000

C) 42000

D) 47000

Ans: 65000

Solutions : Out of 26 alphabets two distinct letters can be chosen in 26p2 ways. Coming to numbers part, there are 10 ways (any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit. Hence there are totally 10X10 = 100 ways. Combined with letters there are 6p2 X 100 ways = 65000 ways to choose vehicle numbers.

2. If the letters of the word SACHIN are arranged in all possible ways and these

Words are written out as in dictionary, then the word ‘SACHIN’ appears at serial

Number?

( a ) 601 ( b ) 600 ( c ) 603 ( d ) 602

Ans: a)

Solutions: if the word started with the letter A then the remaining 5 positions can be filled in `

5! Ways

If it started with c then the remaining 5 positions can be filled in 5! Ways

Similarly if it started with H,I,N the remaining 5 positions can be filled in 5! Ways.

If it started with S then the remaining position can be filled with A,C,H,I,N in

alphabetical order as on dictionary

The required word SACHIN can be obtained after the 5X5!=600 Ways

i.e. SACHIN is the 601th letter

3. If repetition of the digits is allowed, then the number of even natural numbers

Having three digits is

( a ) 250 ( b ) 350 ( c ) 450 ( d ) 550

Ans: c)

Solution:In a 3 digit number one’s place can be filled in 5 different ways with (0,2,4,6,8)

10’s place can be filled in 10 different ways

100’s place can be filled in 9 different ways

There fore total number of ways = 5X10X9 = 450

4. Determine the total number of five-card hands that can be drawn from a deck of 52 cards.

A) 2598960 B)2596800 C)2986800 D)2598708

Answer: A

Solution:

When a hand of cards is dealt, the order of the cards does not matter. If you are dealt

two kings, it does not matter if the two kings came with the first two cards or the last two

cards. Thus cards are combinations. There are 52 cards in a deck and we want to

know how many different ways we can put them in groups of five at a time when order

does not matter. The combination formula is used.

C(52,5) = 2,598,960

Therefore there are 2,598,960 different ways to create a five-card hand from a deck of 52

cards.

5. A school has scheduled three volleyball games, two soccer games, and four

basketball games. You have a ticket allowing you to attend three of the games. In how

many ways can you go to two basketball games and one of the other events?

A)6 B)5 C)30 D)11

Answer: C

Explanation :

Since order does not matter it is a combination.

The word AND means multiply.

Given 4 basketball, 3 volleyball, 2 soccer.

We want 2 basketball games and 1 other event. There are 5 choices left.

C(n,r)

C(How many do you have, How many do you want)

C(have 4 basketball, want 2 basketball)*C(have 5 choices left, want 1)

C(4,2)*C(5,1)

(6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and one

of the other events.

6. A local delivery company has three packages to deliver to three different homes.

if the packages are delivered at random to the three houses, how many ways are there

for at least one house to get the wrong package?

A) 3 B) 5 C) 3! D) 5!

Answer : B

Solution:

The possible outcomes that satisfy the condition of “at least one house gets the wrong

package” are:

One house gets the wrong package or two houses get the wrong package or three houses get

the wrong package.

We can calculate each of these cases and then add them together, or approach this problem

from a different angle.

The only case which is left out of the condition is the case where no wrong packages are

delivered.

If we determine the total number of ways the three packages can be delivered and then

subtract the one case from it, the remainer will be the three cases above.

There is only one way for no wrong packages delivered to occur. This is the same as

everyone gets the right package. The first person must get the correct package and the second

person must get the correct package and the third person must get the correct package.

1*1*1 = 1

Determine the total number of ways the three packages can be delivered.

3*2*1 = 6

The number of ways at least one house gets the wrong package is:

6 – 1 = 5

7. How many different four letter words can be formed (the words need not be meaningful)

using the letters of the word “MEDITERRANEAN” such that the first letter is E and the

last letter is R?

A) 59

B) 56

C) 23

D) 11!

2x2x2x3!

Answer : A

Explanatory Answer

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters.

Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the

letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from

the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options – the three can be either Ns or

Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

8. How many lines can you draw using 3 non collinear (not in a single line) points A, B and

C on a plane?

A) 6 B) 4 C) 3 D)2

Solution: C

You need two points to draw a line. The order is not important. Line AB is the same as line

BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did

with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and

CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that’s because in this problem the order of the

points A, B and C is not important.

9. How many ways can 10 letters be posted in 5 post boxes, if each of the post boxes can

take more than 10 letters?

(A)5^10 (B)10^5 (C)10P5 (D) 10C5

Correct Answer A

Solution:

Each of the 10 letters can be posted in any of the 5 boxes.

So, the first letter has 5 options, so does the second letter and so on and so forth for all of the

10 letters.

i.e. 5*5*5*….*5 (upto 10 times)

= 5 ^ 10.

10. How many number of times will the digit ‘7′ be written when listing the integers from 1

to 1000?

(A)271 (B)300 (C)252 (D) 304

Correct Answer B

Solution:

7 does not occur in 1000. So we have to count the number of times it appears between 1

and 999. Any number between 1 and 999 can be expressed in the form of xyz where

0 < x, y, z < 9.

1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc

This means that 7 is one of the digits and the remaining two digits will be any of the other

9 digits (i.e 0 to 9 with the exception of 7)

You have 1*9*9 = 81 such numbers. However, 7 could appear as the first or the second

or the third digit. Therefore, there will be 3*81 = 243 numbers (1-digit, 2-digits and

3- digits) in which 7 will appear only once.

In each of these numbers, 7 is written once. Therefore, 243 times.

2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77

In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with

the exception of 7).

There will be 9 such numbers. However, this digit which is not 7 can appear in the first or

m second or the third place. So there are 3 * 9 = 27 such numbers.

In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.

3. The number in which 7 appears thrice – 777 – 1 number. 7 is written thrice in it.

Therefore, the total number of times the digit 7 is written between 1 and 999 is

243 + 54 + 3 = 300

11. There are 2 brothers among a group of 20 persons. In how many ways can the group be

arranged around a circle so that there is exactly one person between the two brothers?

(A)2*17! (B) 18!*18 (C) 19!*18 (D)2*18! (E)2*17!*17!

Correct Answer D

Solution

Circular Permutation ‘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 =2/(n – 1)!

Let there be exactly one person between the two brothers as stated in the question.

If we consider the two brothers and the person in between the brothers as a block, then there will

17 others and this block of three people to be arranged around a circle.

The number of ways of arranging 18 objects around a circle is in 17! ways.

Now the brothers can be arranged on either side of the person who is in between the brothers in

2! ways.

The person who sits in between the two brothers could be any of the 18 in the group and can be

selected in 18 ways.

Therefore, the total number of ways 18 * 17! * 2 = 2 * 18!.

12. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is

empty, if any of the boxes may hold all of the toys?

(A)20(B)30(C)25(D)600(E)480

Correct Answer is 25 – Choice (C)

Explanation:

The toys are different; The boxes are identical

If none of the boxes is to remain empty, then we can pack the toys in one of the following ways

a. 2, 2, 1

b. 3, 1, 1

Case a. Number of ways of achieving the first option 2 – 2 – 1

Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be

selected in 3C2 ways and the last toy can be selected in 1C1 way.

However, as the boxes are identical, the two different ways of selecting which box holds the first

two toys and which one holds the second set of two toys will look the same. Hence, we need to

divide the result by 2.

Therefore, total number of ways of achieving the 2 – 2 – 1 option is ways:

5C2*3C2 = 10*3=15WAYS

2 2

Case b. Number of ways of achieving the second option 3 – 1 – 1

Three toys out of the 5 can be selected in 5C3 ways. As the boxes are identical, the remaining

two toys can go into the two identical looking boxes in only one way.

Therefore, total number of ways of getting the 3 – 1 – 1 option is 5C3 = 10 = 10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes

= number of ways of achieving Case a + number of ways of achieving Case b

= 15 + 10 = 25 ways.

13. How many necklace of 12 beads each can be made from 18 beads of different colours?

A) B) C) D) E)

Ans. Here clock-wise and anti-clockwise arrangements are same.

Hence total number of circular–permutations: 18P12/2×12 = 18!/(6 x 24)

14. Goldenrod and No Hope are in a horse race with 6 contestants. How many different

arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of

the horses finish the race?

(A) 720(B) 360(C) 120(D) 24(E) 21

Answer (B)

Solution:

two horses A and B, in a race of 6 horses… A has to finish before B

if A finishes 1… B could be in any of other 5 positions in 5 ways and other horses finish

in 4! Ways, so total ways 5*4!

if A finishes 2… B could be in any of the last 4 positions in 4 ways. But the other

positions could be filled in 4! ways, so the total ways 4*4!

if A finishes 3rd… B could be in any of last 3 positions in 3 ways, but the other positions

could be filled in 4! ways, so total ways 3*4!

if A finishes 4th… B could be in any of last 2 positions in 2 ways, but the other positions

could be filled in 4! ways, so total ways… 2 * 4!

if A finishes 5th .. B has to be 6th and the top 4 positions could be filled in 4! ways..

A cannot finish 6th, since he has to be ahead of B

therefore total number of ways

5*4! + 4*4! + 3*4! + 2*4! + 4! = 120 + 96 + 72 + 48 + 24 = 360

15. There are 7 non-collinear points. How many triangles can be drawn by joining these

points?

A) 35 B)10 C)8 D)7

Solution: A,

A triangle is formed by joining any three non-collinear points in pairs.

There are 7 non-collinear points

The number of triangles formed = 7C3=35

16. There are 6 bowlers and 9 batsmen in a cricket club. In how many ways can

a team of 11 be selected so that the team contains at least 4 bowlers?

A)126 B)504 C)540 D)1170

Answer:(D)

Possibilities Bowlers Batsmen Number of ways

6 9

1 4 7 6C4 x 9C7

2 5 6 6C5 x 9C6

3 6 5 6C6 x 9C5

6C4 x 9C7 = 15 x 36 = 540

6C5 x 9C6 = 6 x 84 = 504

6C6 x 9C5 = 1 x 126 = 126

Total = 1170

17. If 6Pr = 360 and 6Cr = 15 find r ?

A)4 B)3 C)6 D)5

Answer:(A)

Explanation: nPr = nCr x r !

6Pr = 15 x r!

360 = 15 x r!

r! = 360/15

= 24

= 4 x 3 x 2 x 1

r! = 4!

Therefore r = 4

18. When four fair dice are rolled simultaneously, in how many outcomes will at least one of the

diceshow3?

(1)155 (2)620 (3)671 (4) 625

Correct Answer – (3)

Solution:

When 4 dice are rolled simultaneously, there will be a total of 6^4 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5^4 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

19. In how many different ways can the letters of the word ‘CORPORATION’ be arranged so that the vowels always come together?

A) 810 B) 1440 C) 2880 D) 50400 E)5760

Answer: Option D

Explanation:

In the word ‘CORPORATION’, we treat the vowels OOAIO as one letter.

Thus, we have CRPRTN (OOAIO).

This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.

Number of ways arranging these letters =7! / 2! = 2520

Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in

5!/3!= 20ways.

Required number of ways = (2520 x 20) = 50400.

20. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be

drawn from the box, if at least one black ball is to be included in the draw?

A) 32 B) 48 C) 64 D) 96 E) None of these

Answer: Option C

Explanation:

We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).

Therefore required ways = (3C1*6C2) + (3C2*6C1) + (3C3)

= 3*6*5 + 3*2*6 +1

2*1 2*1

= (45 + 18 + 1)

= 64

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