# 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

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)

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.

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

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

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

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

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