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We have to choose 11 players for a crick...

We have to choose 11 players for a cricket team from 8 batman , 6 bowlers , 4 all rounders and 2 wicket keepers in the following conditions .
The number of selections when almost one all rounder and one wicket keeper will play -

A

`.^(4)C_(1)xx.^(14)C_(11)+.^(2)C_(1)xx.^(14)C_(10)+.^(4)C_(1)xx.^(2)C_(1)xx.^(14)C_(9)+.^(14)C_(11)`

B

`.^(4)C_(1)xx.^(15)C_(11)+.^(15)C_(11)`

C

`.^(4)C_(1)xx.^(15)C_(10)+.^(15)C_(11)`

D

none of these

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The correct Answer is:
A
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The numbers 2,3,4,5 and 6 written on five pieces of paper. Theses pieces of paper are put in a bag and then mixed thoroughly. A man draws at random two pieces of paper from the bag one after the other without replacement. Describe the following events associated with this random experiment : A : the number on the first piece is less than the one on the second piece, B: the sum of the numbers on the two pieces is 8 or 9 , C: the number on the first piece is odd and D: the number on the second piece is greater than 4. Is there any pair of events which are mutually exclusive?

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Knowledge Check

  • We have to choose 11 players for a cricket team from 8 batman , 6 bowlers , 4 all rounders and 2 wicket keepers in the following conditions . Number of selections when a particular batsman and a particular wicket keeper do not want to play together -

    A
    `2.^(18)C_(10)`
    B
    `.^(19)C_(11)+.^(18)C_(10)`
    C
    `.^(19)C_(10)+.^(19)C_(11)`
    D
    none of these

  • We have to choose 11 players for a cricket team from 8 batman , 6 bowlers , 4 all rounders and 2 wicket keepers in the following conditions . Numer of selections when two particular batsmen do not want to play when a paritcular bowler will play -

    A
    `.^(17)C_(10)+.^(19)C_(11)`
    B
    `.^(17)C_(10)+.^(19)C_(11)+.^(17)C_(11)`
    C
    `.^(17)C_(10)+.^(20)C_(11)`
    D
    `.^(19)C_(10)+.^(19)C_(11)`

  • H, He^(+), Li^(2+) are examples of atoms or ions with one electron each . The energy of such atoms when in the n-th energy state (according to Bohr,s theory , n=1,2,3…. =principal quantum number ) is E_n =(-13.6 Z^2)/(n^2) eV (1 eV =1.6xx10^(-19)J) . For the ground state ,n=1 . in order to raise the atom from the ground state to n=f , the suitable incident light should have a wavelength given by lambda=(hc)/(E_f-E_1) . But the atom cannot stay permanently in the f-energy state, ultimately , it comes to the ground state by radiating the extra energy , E_f-E_1 as electromagnetic radiation . The electron of the atom comes from n=f to n=1 in one or more steps using the permitted energy levels . As a result there is a possibility of emission of radiation with more than one wavelength from the atom. Planck's constant =6.63 xx10^(-34)J*s and velocity of light c=3xx10^(8)m*s^(-1) . The wavelength of radiation emitted for the transition of the electron of He^+ ion from n=4 to n=2 is

    A
    952 Å
    B
    975 Å
    C
    1027 Å
    D
    1219 Å

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