## Elitmus Previous Years Questions on Permutations and Combination

# Permutations And Combinations 01

### Permutation And Combinations 02

1.

Greatest five digit number : 43210

Smallest five digit number : 10234

Difference = 43210-10234= 32976

(c)

2. We consider vice-chairman and the chairman as 1 Unit. Now, 9 persons can be arranged along a circular table in 8! ways. And vice-chairman and chairman can be arranged in 2 different ways. Hence required number of ways=2 x 8!

(b)

3. 3 Girls can be selected out of 5 girls in 5C3 ways.

Since number of boys to be invited is not given, hence out of 4 boys, he can invite them (2)4 ways.

Hence required number of ways is = 5C3 x (2)4 = 160

(b)

4.Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls=25=32.

Two adjacent boxes with blue can be got in 4 ways.

i.e. (1, 2) (2, 3) (3, 4) (4, 5)

Three adjacent boxes with blue can be got in 3 ways

i.e. (1, 2, 3), (2, 3, 4) and (3, 4, 5)

Four adjacent boxes with blue can be be got in 2 ways.

i.e. (1234) , (2345)

and five boxes with blue can be got in 1 way.

Hence total number of ways of filling the boxes such that adjacent boxes have blue

(4 + 3 + 2 + 1)=10

(d)

5. To construct 2 roads, three towns can be selected out of 4 in 4 x 3 x 2 = 24 ways. Now, if the third road goes from the third town to the first town, a triangle is formed and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.

(d)

6. Total number of 4 digit numbers that can be formed = 4!. If the number is divisible by 25, then the last two digit are 25. So the first two digits can be arranged in 2! ways.

Hence required probability = 2!/4! = 1/12

(a)

7. Keeping one digit in fixed position, other four can be arranged in 4! ways= 24 ways. Thus each of the 5 digits will occur in each of the five place 4! times. Hence the sum of digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600. So, the sum of all numbers

=600(1+10+100+1000+10000)=6666600

(a)

8. Required number of matches played will be (139 – 1)=138

(c)

9. Required probability is = 1/6

(c)

10. Let the number of students in the front row be x and the number of rows be n.

Hence, number of students in the next rows would be (x – 3) , (x – 6), (x – 9),……. and so on.

Now we have to check for each value of n=3, 4, 5, 6

Firstly take n=3

x + (x – 3) + (x – 6) =630

» 3x = 639 i.e. x = 213 ( Thus, n=3 is possible)

Likewise if n=4

x + (x – 3) + (x – 6)+(x – 9) =630

» 4x – 18 = 630 i.e. x = 162 ( Thus, n=4 is possible)

Likewise if n=5

x + (x – 3) + (x – 6)+(x – 9) +( x -12) =630

» 5x – 30 = 630 i.e. x = 132 ( Thus, n=5 is possible)

Likewise if n=6

x + (x – 3) + (x – 6)+(x – 9)+(x -12)+(x – 15)=630

» 6x – 45 = 630 i.e. x =112.5 (NOT AN INTEGER) ( Thus, n=6 is not possible)

(d)

Set 2 Detailed Solutions

1. Let there be m boys and n girls.

Then nC2 = 45 » n(n – 1)=90» n =10

mC2 = 190 » m(m – 1) = 380 » m=20

Number of games played between one boy and one girl

= 10C1 x 20C1= 10 x 20=200

(a)

2. Any of the 4 colours can be chosen for the first stripe. Any of the remaining 3 colours can be used for the second stripe. The stripe can again be colored in 3 ways. ( We can repeat the colour of the first stripe, but not use the colour of the second stripe).

Similarly, there are 3 ways to colour each of the remaining stripes.

» The number of ways the flag can be coloured is

4 x (3)5 = (12)(3)4

(a)

3. The available digits are 0, 1, 2 …………… 9. The first digit can be chosen in 9 ways ( 0 not acceptable ), the second digit can be accepted in 9 ways ( digits repetition not allowed). Thus the code can be made in 9 x 9=81 ways.

Now, there are only 4 digits which can create confusion 1, 6, 8, 9. The same can be given in the following ways.

Total number of ways confusion can arise = 4 x 3=12

Thus, required answer= 81-12=69

(d)

4.

Number of ways for selecting single digit=2

Number of ways for selecting two digit= 2 x 3=6

Number of ways for selecting three digit= 2 x 3 x 3=18

Number of ways for selecting four digit= 2 x 3 x 3 x 3=54

Number of ways for selecting five digit=2 x 3 x 3 x 3 x 3=162

Number of ways for selecting six digit= 2 x 3 x 3 x 3 x 3 x 3=486

Hence, total number of ways = (2 + 6 + 18 + 54 + 162 + 486)=728

(d)

5. There are 32 black and 32 white square on a chess-board then number of ways in choosing one white and one black square on the chess.

32C1 x 32C1= 32 x 32=1024

Number of ways in which square lies in the same row

White square=4

Black square=4

Number of rows=8

4C1 x 4C1 x 8 = 128

»Number of ways in which square lies in the same column= 128

Total number in which square lie on the same row or same column= 128 + 128=256.

(d)

6. Ist place of the four letter password can be filled in 11 ways.

IInd place of four letter password can be filled in 10 ways.

IIIrd place of four letter password can be filled in 9 ways.

IVth place of four letter password can be filled in 8 ways.

Hence, required number of ways= 11 x 10 x 9 x 8=7920 ways

(a)

7. Three letter password from 26 letters can be selected in 26 x 25 x 24 ways. Three letter password from 15 asymmetric letters can be selected in 15 x 14 x 13 ways.

Hence, three letter password with at least one symmetric letter can be made in (26 x 25 x 24)-(15 x 14 x 13)=12870 ways.

(c)

8. At lease one candidate out of (2n + 1) candidates can be selected in (2n+1 – 1) ways.

» 22n+1 – 1 = 63 » 22n+1 = 64 = (2)6» n=2.5

Since n cannot be a fraction. Hence n=3.

(a)

9. Required number of triangles formed

10C2 x 11 + 11C2 x 10= 45 x 11 + 55 x 10 = 1045

(c)

10. The digit in the unit’s place should be greater than that in the ten’s place. Hence, if digit 5 occupies the unit place then remaining four digits need not to follow any order.

Hence required number of ways = 4!

However, if digit 4 occupies the unit place then 5 cannot occupies the ten’s positions. Hence, digits at the ten’s place will be one among 1, 2 or 3. This can happen in 3 ways. The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence in all we have (3 x 3!) numbers ending in 4.

Similarly, if we have 3 in the unit’s place, the ten’s place can be either 1 or 2. This can happen in 2 ways. The remaining 3 digits can be arranged in the remaining 3 places in 3! ways. Hence we will have (2 x 3!) numbers ending in 3. Similarly, we can find that there will be 3! numbers ending in 2 and no number ending with 1. Hence total number of numbers

= 4! + (3) x 3! + (2)3! + 3!

= 4! + 6 x 3! = 24 + (6 x 6) = 60

(b)