7 - Permutation and Combination Questions Answers

what will be the rank of word OPTION in the dictionary of words formed by letters of the word OPTION

Asked By: JIGYASA KUMAR
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Joshi sir comment

OPTION

ordered way INOOPT

total words = 6!/2! = 360

so rank = (360/6)*2+([60*2]/5)*3+(24/4)*3+(6/3)*0+(2/2)*1+1 = 120+72+18+0+1+1 = 212

 

rank of the number 3241?(digits cannot be repeated)

Asked By: HARSHITH
is this question helpfull: 7 2 read solutions ( 1 ) | submit your answer
Joshi sir comment

total numbers = 4*3*2*1 = 24

on arranging these numbers in order we get 1234

so rank = [24/4]*2+[6/3]*1+[2/2]*1+1 = 16

 

If the letters of the word are arranged as in dictionary, find the rank of the word INDIA.

Asked By: UMESH
is this question helpfull: 4 5 submit your answer
Joshi sir comment

On arranging the letters in ascending order we will get

A D I I N

total no of words = 5*4*3*2*1/2*1 = 60

rank = (60/5)*2 + (24/4)*3 + (6/3)*1 + (2/2)*1 + 1 = 24+18+2+1+1 = 46

here 60/5 means total words/ total letters and 2 is here because in INDIA, I is first letters and this comes at 3rd and 4th place in ascending order arrangement so total letters before it are 2. similarly 24 stands for total words after removing I from the spelling of INDIA and etc 

Let a number n have its base -7 representation as n= 1223334444. the highest  power of 7, that divides the number n! is 

Asked By: AMIT DAS
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Joshi sir comment

 

n = (1223334444)= 1*7+ 2*78 + 2*77 + 3*76 + 3*75 + 3*74 + 4*73 + 4*72 + 4*71 + 4*70

so if we will divide n! by 7x we will get 1*78 + 2*77 + 2*76 + 3*75 + 3*74 + 3*73 + 4*72 + 4*71 + 4*70 as x

 

 

A light example

if we divide (1*72 + 2*71 + 3*70)! by 7x we get x = 1*71 + 2*70 = 9     

 

Let α, β and γ be the cube roots of p (p < 0). For any real numbers x, y and z, the value of  (xα+yβ+zγ) (xβ+yγ+zα) is

Asked By: HIMANSHU MITTAL
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Joshi sir comment

According to the given condition p is a negative number

consider p = -k so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω2

On applying given values you will get the answer as  ω or ω2

you should use ω3 = 1

    

Solution by Joshi sir

 

According to the given condition p is a negative number

consider p = -k = (k1/3)3(-1) so cube root of p will be α = -k1/3, β = -k1/3ω, γ = -k1/3ω, here -1, -ω, -ω2 are the cube roots of -1

On applying given values on the given expression, you will get  

x(-k1/3)+y(-k1/3ω)+z(-k1/3ω2)/x(-k1/3ω)+y(-k1/3ω2)+z(-k1/3)

=x+yω+zω2/xω+yω2+z

first multiply ω in numerator and denominator we get answer as 1/ω = ω2

second interchange the value of α, β, γ for getting answer as ω

you should use ω3 = 1

    

If A and B are two independent events such that and P(A'∩B)=2/7 and P(A∩B')=1/5, then P(A) is

Asked By: HIMANSHU MITTAL
is this question helpfull: 6 1 submit your answer
Joshi sir comment

 

Let p(A) = x and p(B) = y

we know that p(A∩B) + p(A∩B') = p(A)

and                  p(A∩B) + p(B∩A') = p(B)

and for independent event p(A∩B) = p(A)p(B)

so on applying all the given conditions relation etween x and y will be

xy + 1/5 = x

xy + 2/7 = y

solve now

Total number of non-negative integral solutions of 2x + y + z = 21 is

Asked By: HIMANSHU MITTAL
is this question helpfull: 8 1 read solutions ( 1 ) | submit your answer
Joshi sir comment

 

You can solve this question by the following method

possile algebric expression for the given equation is

(x0+x2+x4+........+x20)(x0+x1+x2+..................+x21)(x0+x1+x2+x3+....................x21)                 (i)

Here three brackets are for x , y and z, and powers are based on the range of least to greatest possile values of x, y, z

on solving eq (i) will become (x0+x2+x4+...........+x20)(1-x22)2(1-x)-2

now we have to calculate coefficient of x21 in this expression so (1-x22)2 can be omitted

general term of (1-x)-2 is (r+1)xr

so required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132

Solution by Joshi sir

 

You can solve this question by the following method

possile algebric expression for the given equation is

(x0+x2+x4+........+x20)(x0+x1+x2+..................+x21)(x0+x1+x2+x3+....................x21)                 (i)

Here three brackets are for x , y and z, and powers are based on the range of least to greatest possile values of x, y, z

on solving eq (i) will become (x0+x2+x4+...........+x20)(1-x22)2(1-x)-2

now we have to calculate coefficient of x21 in this expression so (1-x22)2 can be omitted

general term of (1-x)-2 is (r+1)xr

so required coefficient = 22+20+18+16+14+12+10+8+6+4+2 = 132

 

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