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
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)
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.
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
n = (1223334444)7 = 1*79 + 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
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
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 = -k1/3ω2 , 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
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
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
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