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A train goin from delhi to jaipur stops at 7 intermediate stations. 5 persons enter the train during the journey with 5 difefrent tickets of same class . How mant different set of tickets they could had??Solution) This is the condition if they all board the train from Delhi, though if they can board the train from any station . so here is the solutionEvery person can have two choices 1) Boarding Station 2)Destinationif B.S= Delhi Des=1,2,3,4,5,6,7,Jaipur = 8 choicesif B.S.= 1 Des=2,3,4,5,6,7,Jaipur = 7 choicesif B.S.= 2 Des=3,4,5,6,7,Jaipur = 6 choices...therefore the total type of tickets that a person can have =8+7+6+5+4+3+2+1 = 36choicesLikewise all the fove people can have these choices (May be recurring) .Then the total choices may be 365Though if they have different tickets then the permutation possible are 36*35*34*33*32
Hi, this is EBADULLA its about math assignment. 1 application of complex analysis used in thermodynamics. . what all uses are there in that... plz let mee know this answer.
Q. Illustrate Pythagorean Theorem? Ans. You have definitely seen the Pythagorean Theorem before, so a 2 + b 2 = c 2 should look familiar to you. The Pythagorean Theor
The price of gasoline is $1.349 cents per gallon. If the price increases through three tenths of a cent, what will the price of gasoline be? Three tenths of a cent can be writt
verify 4(sin^4 30^0+cos60^0 )-3(cos^2 ?45?^0-sin^2 90^0 )=2
Solve the subsequent IVP. y′′ + 11y′ + 24 y = 0 y (0) =0 y′ (0)=-7 Solution The characteristic equation is as r 2 +11r + 24 = 0 ( r + 8) ( r + 3) = 0
1. Show that there do not exist integers x and y for which 110x + 315y = 12. 2. If a and b are odd integers, prove that a 2 +b 2 is divisible by 2 but is NOT divisible by 4. H
Hi may i know how to substract the (ID)colum matrix from (K)square matrix as per equation below. E = (K - ID)^-1 S K is m*m matrix I is idntity matrix d is column vector s is col
HOW CAN WE TAKE SUPPOSE THE VALUES OF X AND Y
Product and Quotient Rule : Firstly let's see why we have to be careful with products & quotients. Assume that we have the two functions f ( x ) = x 3 and g ( x ) = x 6 .
how to solve for x
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