Practice question on permutation and combination 31-45 (Set 3)

permutation and combination

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Question no. 31

In a plane , there are 10 points out of which 4 are collinear , then the number of triangles that can be formed by joining these points are

looks_one 60

looks_two 116

looks_3 120

looks_4 None of these

Option looks_two 116

Solution :
No. of triangles = ⁿC₃ - ^pC₃
=¹⁰C₃ - ⁴C₃ = 120 - 4 = 116

Question no. 32

The number of straight lines joining 8 points on a circle is

looks_one 8

looks_two 16

looks_3 24

looks_4 28

Option looks_4 28

Solution :
No. of straight line = n(n-1)/2 = 8 *7 / 2 = 28

Question no. 33

The number of diagonals in a polygon of m sides is

looks_one m(m-5)/2!

looks_two m(m-1)/2!

looks_3 m(m-3)/2!

looks_4 m(m-2)/2!

Option looks_3 m(m-3)/2!

Solution :
The number of diagonals in a polygon of m sides is m(m-3)/2!

Question no. 34

How many triangles can be drawn by means of 9 non collinear points

looks_one 84

looks_two 72

looks_3 144

looks_4 126

Option looks_one 84

Solution :
No of ways = ⁿC₃ = ⁹C₃ = 9!/(6!3!) = 84

Question no. 35

How many triangles can be formed bt joining four points on a circle

looks_one 4

looks_two 6

looks_3 8

looks_4 10

Option looks_one 4

Solution :
No. of triangles = ⁿC₃ = ⁴C₃ = 4

Question no. 36

The number of diagonals in a octogen will be

looks_one 28

looks_two 20

looks_3 10

looks_4 16

Option looks_two 20

Solution:
No .of diagonal = m(m-3)/2 = 8*5/2 = 20

Question no. 37

The number of triangles that can be formed by 5 points in a line and 3 points on a parallel line is

looks_one⁸C₃

looks_two⁸C₃ - ⁵C₃

looks_3⁸C₃ - ⁵C₃ -1

looks_4 None of these

Option looks_3⁸C₃ - ⁵C₃ -1

Solution :
No. of triangles =⁸C₃ - ⁵C₃ - ³C₃
= ⁸C₃ - ⁵C₃ -1

Question no. 38

Out of 10 white , 9 blacks and 7 red balls , the numbers of ways in which selection of one or more balls can be made , is

looks_one 881

looks_two 891

looks_3 879

looks_4 892

Option looks_3 879

Solution :
No. of ways = (p+1)(q+1)(r+1) -1 = 11*10*8 - 1 = 880 - 1 = 879

Question no. 39

In how many ways a team of 11 players can be formed out of 25 players , if 6 out of them are always be included and 5 are always to be excluded

looks_one 2020

looks_two 2002

looks_3 2008

looks_4 8002

Option looks_two 2002

Solution :
out of 25 , 5 are excluded and 6 are always included , So we can choose 5 players from 25-11 (14)
No. of ways = ¹⁴C₅ = 2002

Question no. 40

If ⁸C_r = ⁸C_r+2 then the value of ^rC₂ is

looks_one 8

looks_two 3

looks_3 5

looks_4 2

Option looks_two 3

Solution :

Question no. 41

ⁿC_r + 2.ⁿC_r-1 + ⁿC_r-2 =

looks_one ⁿ⁺¹C_r

looks_two ⁿ⁺¹C_r+1

looks_3 ⁿ⁺²C_r

looks_4 ⁿ⁺²C_r+1

Option looks_3 ⁿ⁺²C_r

Solution :
ⁿC_r + 2.ⁿC_r-1 + ⁿC_r-2

Question no. 42

ⁿC_r:ⁿC_r-1

looks_one (n-r)/r

looks_two (n+r-1)/r

looks_3 (n-r+1)/r

looks_4 (n-r-1)/r

Option looks_3 (n-r+1)/r

Solution :

Question no. 43

How many numbers of 6 digits can be formed from the digits of the number 112233

looks_one 30

looks_two 60

looks_3 90

looks_4 120

Option looks_3 90

No. of ways = 6!/(2!2!2!) = 30 * 3 = 90

Question no. 44

If n and r are two positive integers such that n ≥ r , then ⁿC_r-1 + ⁿC_r =

looks_one ⁿC_n-r

looks_two ⁿC_r

looks_3 ^n-1C_r

looks_4 ⁿ⁺¹C_r

Option looks_4 ⁿ⁺¹C_r

Solution :

Question no. 45

The least value of natyural number n satisfying C(n,5) + C(n,6) > C(n+1 , 5) is

looks_one 11

looks_two 10

looks_3 12

looks_4 13

Option looks_one 11

Solution :

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