# 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**

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**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**

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**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)_{7 }= 1*7^{9 }+ 2*7^{8} + 2*7^{7} + 3*7^{6} + 3*7^{5} + 3*7^{4} + 4*7^{3} + 4*7^{2} + 4*7^{1} + 4*7^{0}

so if we will divide n! by 7^{x} we will get 1*7^{8}^{ }+ 2*7^{7} + 2*7^{6} + 3*7^{5} + 3*7^{4} + 3*7^{3} + 4*7^{2} + 4*7^{1} + 4*7^{0} as x

A light example

if we divide (1*7^{2} + 2*7^{1} + 3*7^{0})! by 7^{x} we get x = 1*7^{1 }+ 2*7^{0} = 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 = -k^{1/3}, β = -k^{1/3}ω, γ = -k^{1/3}ω^{2}

On applying given values you will get the answer as ^{ }ω or ω^{2}

you should use ω^{3} = 1

^{ }

**Solution by Joshi sir**

According to the given condition p is a negative number

consider p = -k = (k^{1/3})^{3}(-1) so cube root of p will be α = -k^{1/3}, β = -k^{1/3}ω, γ = -k^{1/3}ω^{2 }, here -1, -ω, -ω^{2} are the cube roots of -1

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

x(-k^{1/3})+y(-k^{1/3}ω)+z(-k^{1/3}ω^{2})/x(-k^{1/3}ω)+y(-k^{1/3}ω^{2})+z(-k^{1/3})

=x+yω+zω^{2}/xω+yω^{2}+z

first multiply ω in numerator and denominator we get answer as 1/ω = ω^{2}

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

you should use ω^{3} = 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**

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**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 2*x* + *y* + *z* = 21 is

**Asked By: HIMANSHU MITTAL**

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**Joshi sir comment**

You can solve this question by the following method

possile algebric expression for the given equation is

(x^{0}+x^{2}+x^{4}+........+x^{2}^{0})(x^{0}+x^{1}+x^{2}+..................+x^{21})(x^{0}+x^{1}+x^{2}+x^{3}+....................x^{21}) (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 (x^{0}+x^{2}+x^{4}+...........+x^{2}^{0})(1-x^{22})^{2}(1-x)^{-2}

now we have to calculate coefficient of x^{21 }in this expression so (1-x^{22})^{2} can be omitted

general term of (1-x)^{-2 }is (r+1)x^{r}

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

(x^{0}+x^{2}+x^{4}+........+x^{2}^{0})(x^{0}+x^{1}+x^{2}+..................+x^{21})(x^{0}+x^{1}+x^{2}+x^{3}+....................x^{21}) (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 (x^{0}+x^{2}+x^{4}+...........+x^{2}^{0})(1-x^{22})^{2}(1-x)^{-2}

now we have to calculate coefficient of x^{21 }in this expression so (1-x^{22})^{2} can be omitted

general term of (1-x)^{-2 }is (r+1)x^{r}

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