7

Q1. How many 6-digits numbers can be formed by using digits 5, 5, 6, 8, 8, 9?

• 6!
• 180
• 3!
• 120

Solution

We have 6 digits out of which  2 are of one kind and other 2 are of second kind So, the number of permutations is given by = 180

Q2. A room has 8 doors. In how many ways, a man can enter in the room through one door and exit through a different door?

• 56
• 8
• 5040
• 40320

Solution

A person can enter in a room through any one of 8 doors and exit from a different door in 7 ways (remaining 7 doors). Total number of ways = 8 x 7 = 56

Q3. ⁸C_r = ⁸C_p. So

• r = p or  r = 2p
• r = p or r + p = 8
• r = p or r - p = 8
• r = p or r = 8p

Solution

ⁿC_x = ⁿC_y, then either x = y or x + y = n So, using this property, we get the answer.

Q4. What is the number of diagonals that can be drawn by joining the vertices of a hexagon?

• 12
• 9
• 14
• 5

Solution

A line is obtained by joining two vertices. So out of 6 vertices of the hexagon, the number of line segments = C(6,2) = 15 But out of these 6 are sides, so the remaining must be diagonals. = 15 - 6 = 9

Q5. What is the value of ⁿp_o

• zero
• 1
• n
• n!

Solution

n!/(n - 0)! = n!/n! = 1

Q6. Suppose there are four roads from station A to station B and 3 roads between station B and station C. Find the number of ways in which one can drive from station A to station C via station B.

• 12
• 16
• 7
• 4

Solution

There are four ways in which one can drive from A to B and for each way from A to B, there are three ways to drive from B to C. By FCPP, the total number of ways one can drive from A to C via B is 4 3 = 12

Q7. Four alphabets A, M, P, O are purchased from a warehouse. How many ordered pairs of initials can be formed using these?

• 10
• 6
• 12
• 18

Solution

Required number=P(4,2) = 4!/2! = 12

Q8. How many three digit odd numbers can be formed by using the digits 1,2,3,4,5,6 if the repetition of digit is not allowed:

• 216
• 120
• 60
• None of these

Solution

For a number to be odd, we must have 1,3 or 5 at the unit's place.  So there are 3 ways for filling the unit's place.Since the repetition of digits is not allowed, the ten's place can be filled with any of the remaining 5 digits in 5 ways.  Now, four digits are left,  so hundred's place can be filled in 4 ways.So, required number of numbers = 3 5 4 = 60

Q9. The number of ways in which three different rings can be worn in four fingers with at most one in each finger, are

• 6
• 12
• 24
• 64

Solution

The total number of ways is the arrangement of 4 fingers, taken three at a time. So, number of ways =

Q10. What is the number of 5 digit numbers that can be formed using the digits 2,3,4,5,9 so that no digit is repeated?

• 120
• 240
• 60
• 45

Solution

The given five digits can be formed to make five digit numbers in P(5,5) ways = 5! ways = 120

Q11. How many different words can be formed using the letters of the word BHARAT, which begin with B and end with T?

• 24
• 36
• 16
• 12

Solution

Since the positions of B and T are fixed, we only need to arrange the remaining letters. i.e. 4 letters, of which 2 are of one type and the remaining 2 are of different type each.So the number of words=4!/[2!*1!*1!]=12

Q12. In how many ways can 4 red, 3 yellow and 2 green chairs be arranged in a row if the chairs of the same colour are indistinguishable?

• 362880
• 15120
• 15120
• 1260

Solution

Total number of chairs is (4 + 3 + 2) = 9. Out of 9 chairs, 4 are of the first kind (red), 3 are of the second kind (yellow) and 2 are of the third kind (green). Therefore, the number of arrangements =

Q13. If boys and girls sit alternately then the number of ways in which 3 boys and 3 girls can be seated in a row is

• 36
• 52
• 72
• 144

Solution

Q14. The measure of an interior angle of a regular polygon is 140^o. The number of sides and diagonals in this polygon are:

• 7,14
• 7,17
• 8,20
• 9,27

Solution

Let n be the number of sides in this polygon. Number of diagonals =

Q15. A coin is tossed 6 times, in how many throws can 4 heads and 2 tails be obtained?

• 10
• 15
• 18
• 24

Solution

Total throws = 6Oou of 6, 4 throws should give one type of outcome and 2 of another type.So required number =6!/[4!*2!]=15

Q16. In how many ways can a cricket team be selected out of 15 players if a particular player has to be included?

• C(15,11)
• C(14,11)
• C(14,10)
• C(15,10)

Solution

Since a particular player must be included, so the choice of 10 players is to be made out of 14 players.

Q17. A journey from place A to place B can be done by three different modes, i.e. train, bus or aeroplane. If a person wants to go from A to B and return, in how many ways this can be done?

• 6
• 3
• 9
• 5

Solution

The forward journey can be taken up in 3 ways and same for the return journey. Hence the total number of ways = 3 3 = 9

Q18. A group consists of 4 girls and 7 boys. in how many ways a team of 5 members be selected consisting of 2 girls and 3 boys.

• ¹¹C₅
• 2! x 3!
• 210 ways
• 120 ways

Solution

Q19. The sum of the first 50 natural numbers is

• 1275
• 1300
• 1325
• 1350

Solution

The sum of the first n natural numbers is given by n(n+1)/2 So, substituting n = 50 we get25 51 = 1275.

Q20. If all permutations of the letters of the word APPLE are arranged as in the dictionary, what is the 35^th word?

• PPLEA
• PPAEL
• LPPAE
• LAPPE

Solution

The given word has a total of 5 letters , of which, 2 are of one type. The first word in the dictionary will start with A.The remaining 4 letters P,P,L,E can be arranged among themselves in 4!/2! =12 ways.Then starting with  the next letter E, there will be 4!/2!=12 ways.Next, starting with L, again there will be 4!/2!=12 ways.Thus there will be 36 words  before the words starting with P.the 36^th word will be LPPEA and so the 35^th word will be LPPAE.

Q21. Find the total number of ways in which n distinct objects can be put into two different boxes

• n(n - 1)
• n²
• 2ⁿ
• none of these

Solution

Let the two boxes be b₁ and b₂, we observe that there are two choices for each of the n objects.  Therefore, by fundamental principle of multiplication.Total number of ways = 2 2 2 .......2 = 2ⁿ.

Q22. The number of 4 letters word which can be formed from the letters of the word PIGEON are

• 30
• 120
• 360
• 24

Solution

The number of 4 letters word which can be formed from the letters of the word PIGEON are

Q23. How many number of four digits can be formed with the digits 1,2,3,4,5 if the digit can be repeated in any number of times?

• 625
• 9000
• 5!
• 55

Solution

Since repetition of digits is allowed, therefore each of the four places can be filled in 5 ways (using any of 1,2,3,4,5).Hence, the required number of numbers = 5 5 5 5 5 = 625.

Q24. How many numbers divisible by 5 and lying between 4000 and 5000 can be formed from the digits 4,5,6,7 and 8?

• 5
• 125
• 625
• 25

Solution

Clearly, a number beween 4000 and 5000 must have 4 at thousand's place.  Since the number is divisible by 5 it must have 5 at unit's place.  Now, each of the remaining places (hundred's and ten's) can be filled in 5 ways.Hence, total number of required numbers = 1 5 5 1 = 25.

Q25. If (n + 1)! = 20(n - 1)!, then n is equal to

• 4
• 5
• -5
• 20

Solution

(n + 1)! = 20(n - 1)!(n + 1) n (n - 1)! = 20 (n - 1)!

Q26. In how many ways can a cricket team of 11 players be chosen out from a squad of 14 players, if 5 particular players are always chosen?

• 84
• 144
• 1001
• 1365

Solution

If five particular players are always chosen. This means that 6 players are selected out of remaining 9 players. Required number of ways =

Q27. Sonia has 10 balloons out of which 5 are red, 2 white, 2 blue and 1pink, which she wants to use for the decoration. Her favourite pink colour balloon should be filled with toffees and should be put at the centre of the room above the cake table and remaining 9 at the wall behind the cake table. How many ways she can arrange the balloons?

• 756
• 7560
• 956
• 9!

Solution

Q28. How many words beginning with 'T' and ending with 'E' can be formed using the letters of the word"TRIANGLE" ?

• 24
• 120
• 625
• 720

Solution

The word triangle has 8 different letters. The letter "T" and "E" are fixed in the beginning and at the end, then remaining 6 letters can be arranged in

Q29. C(n,12) = C(n,8)N = ?

• 10
• 12
• 14
• 20

Solution

Q30. P(5, r) = 2P(6,r - 1) R = ?

• 15
• 3
• 12
• 8

Solution

and r=3 but r should be less than 5 so r10 so r=3

Q31. How many three digit odd numbers can be formed by using the digits 5,6,7,8 if the repetition of digits is allowed?

• 64
• 32
• 24
• None of these

Solution

For a number to be odd, we must have 5,7 at the unit's place.  So there are 2 ways of filling the unit's place. Since the repetition of digits is allowed, the ten's place and hundred's place can be filled with any of the 4 digits (5,6,7,8).Hence, the required number of numbers = 2 4 4 = 32.

Q32. 4! + 5! =

• 120
• 144
• 132
• 156

Solution

4! + 5! = 4! [1 + 5]= 4! [6]= 4 x 3 x 2 x 1 [6]= 24 x 6 = 144

Q33. What is the value of ⁿp_n?

• zero
• 1
• n
• n!

Solution

Q34. There are 3 white, 4 red and 1 blue marbles in a bag. They are drawn one by one and arranged in a row. Assuming that all 8 marbles are drawn, determine the number of different arrangements if marbles of same colour are indistinguishable.

• 56
• 240
• 280
• 296

Solution

Q35. If (n + 2)! = 30 n! then n is equal to

• 4
• 7
• -7
• -4

Solution

(n + 2)! = 30 n!(n + 2) (n + 1) n! = 30 n!

Q36. How many triangles can be formed by joining four points on a circle?

• 4
• 2
• 8
• 6

Solution

As the triangles are formed by joining 3 pointsSo number of triangles is=4

Q37. There are 8 doors to a college hall. In how many ways can a student enter through one door and get out through another?

• 64
• 56
• 30
• 12

Solution

The student can get into the hall in 8 ways and go out through any of the doors other than the one entered through i.e. in 7 ways. So using the Fundamental Principle of counting the total number of ways = 8 7 = 56

Q38. Number of signals that can be made using given 4 flags of which 3 are blue and 1 is red?

• 4
• 12
• 4!
• 3!

Solution

We have 4 flags in all and   out of which 3 are alike Here, we are suppose to arrange 4 objects, out of which 3 are of one kind So, the total possible arrangements would be Hence 4 signals can be generated.

Q39. The number of two digit even numbers that can be formed using the digits 3,4,5,6,7 where the digits can be repeated is

• 10
• 15
• 12
• 16

Solution

The unit's place can be filled up only in 2 ways, either 4 or 6 as we have to get only even numbers. Since there's no restriction on repeating numbers, any one of the five numbers out of 3,4,5,6,7 can be used to fill up the ten's place. So there are 5 ways in which the ten's place can be filled up. So by the Fundamental Principle of Counting, the the total number of ways in which this can be done, is, 5 2 = 10 .Hence  the answer.

Q40. In how many ways can two 10-paise coins, two 20-paise coins, three 25-paise coins and one 50-paise coin be distributed among 8 children so that each child gets only one coin?

• 1450
• 1325
• 1680
• 1400

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