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  1. #11
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    Quote Originally Posted by Md. Minuddin View Post
    How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
    A. 6

    B. 7
    C. 5
    D. 8
    E. 18

    Answer-B=7
    a big thank you to u mohammad for all ur answers most of which u get right. but plz post ur explanations with the answrs as i already have the answers. i want to kow how they got the answer.
    the unlimited potential that lies within you. let it slumber no more.

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    Quote Originally Posted by Md. Minuddin View Post
    There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
    A. 5
    B. 21
    C. 33
    D. 60
    E. 6

    Answer is B=21 (6+5+4+3+2+1)

    explanation plz??
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    Quote Originally Posted by Md. Minuddin View Post
    Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

    A. 210

    B. 29

    C. 3*28

    D. 3*29

    E. None of these

    The answer should 29. Is answer B 29 or 29????

    spot on buddy. oa-b. but plz post the explanation. thx in advance.
    the unlimited potential that lies within you. let it slumber no more.

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    Quote Originally Posted by india_chintan View Post
    explanation plz??
    There are 6 boxes numbered 1, 2 ...6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is
    We have to use at least 1 green ball.
    So we can use 1, 2, 3, 4, 5 or 6 green ball.
    If we use 1 green ball then the probable way is 6C1=6
    If we use tow green ball then as the two box will have to be consecutively numbered for this reason we have to take this two ball as a single number and the probable way will be 5C1=5
    and same way if we use 3 green ball then 4C1=4 and so on

    total=6+5+4+3+2+1=21

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    Quote Originally Posted by india_chintan View Post
    spot on buddy. oa-b. but plz post the explanation. thx in advance.
    Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head?

    Just ten coin tossed simultaneously and we have to find the number of way in we can get third coin turn up as head. The rest 9 coin can be turn up either head or tail. If we toss n coin simultaneously then the probable outcome is 2n. So for 9 coin it is 29.
    Example for two coin, HH, HT, TH, TT= 4 = 22
    For three coin, HHH,HHT,HTH,THH, HTT,THT,TTH,TTT=8=23


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    Quote Originally Posted by india_chintan View Post
    a big thank you to u mohammad for all ur answers most of which u get right. but plz post ur explanations with the answrs as i already have the answers. i want to kow how they got the answer.
    How many different positive integers exist between 10^6and 10^7, the sum of whose digits is equal to 2?
    Existing positive integer will be in between 106 =1000000 and 107 =10000000.
    As the sum of the digit is two so the number may start only with 1 or 2.
    If it start with 1 then there will be one more 1 and the rest will be zero.
    As it start with 1 and there are more 6 position the rest 1 may sit in
    6C 1=6 way.
    If it start with 2 then the rest digit will be zero. So here we will get only one number

    Total=6+1 =7

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    Quote Originally Posted by Md. Minuddin View Post
    What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re-arranged?

    A. 1/4


    B. 1/6

    C. 1/3
    D. 1/24
    E. 1/12

    Answer is 1/4
    Answer is 1/4
    There are four letter in math and we can rearrange this four letter in 4! ways.
    And constant remain in order word is 4C1 (math,amth, mtah and mtha) =4
    So probability is 4/4!=4/(4*3*2*1)=1/6



    u said is answer is 1/4 nd caluclated it as 1/6. ??
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    Quote Originally Posted by india_chintan View Post
    u said is answer is 1/4 nd caluclated it as 1/6. ??
    I made mistake to write actual answer should be 1/6.

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    Quote Originally Posted by india_chintan View Post
    understood the 4!*3! i.e. 3! is for vowels and 4! is for all letters. y is it divided by 2!.
    The reason why you divide 2 is because A appears twice. Can you explain why it is 4! * 3!? I am not sure why it should be 4! thanks

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    Quote Originally Posted by lazyjava View Post
    The reason why you divide 2 is because A appears twice. Can you explain why it is 4! * 3!? I am not sure why it should be 4! thanks
    Actually ans can be written 4!*(3!/2!)

    Here 4! is the total way of arrangement taking the vowel as a single letter.

    (3!/2!) is the way of arrangement of the three vowel

    If in a word contain n letter and if among them, there is some letter repeat p, q, ........r times then total way of combination is

    n!/(p!*q!....r!)

    Example: MATHEMATICS

    Here total way of arrangement will be 11!/(2!*2!*2!) as A, M and T repeat 2 times

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