abhishek_mumbai Posted May 14, 2009 Share Posted May 14, 2009 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 E. 1/12 Quote Link to comment Share on other sites More sharing options...
abhishek_mumbai Posted May 14, 2009 Share Posted May 14, 2009 In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? A.6!/2! B. 3!*3! C.4!/2! D.4!*3!/2! E.3!*3!/2! Answer is D.4!*3!/2! (AAu)BCS => can be arranged in 4! ways.. then AAU can be arranged in 3!/2! ways total 4!*3!/2! Quote Link to comment Share on other sites More sharing options...
abhishek_mumbai Posted May 14, 2009 Share Posted May 14, 2009 How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R? A. 59 B.11!/3!*2!*2!*2! C. 56 D. 23 E.11!/2!*2!*2! before using, the number of times alphabet appear is m=1; e=3; r=2; a=2; n=2; d=1; i=1; t=1 We make the word E _ _ R so number of times E and R is reduced by 1 so m=1; e=2; r=1; a=2; n=2; d=1; i=1; t=1 total combinations = 11!/2!*2!*2! Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted May 14, 2009 Share Posted May 14, 2009 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 My answer is (A) 6 110000 101000 100100 100010 100001 200000 106= 1000000 107= 10000000 There is 1000001 1000010 1000100 1001000 1010000 1100000 2000000 Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 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 10^6=1000000 , 10^7= 10000000 any integer between these two values will have 7 digits for a 7 digit number to have 2 as sum of its digits, there are only two possibilities : 1) one of the digits has to be 2 and the rest 0...for the number to remain a 7 digits number, that one digit has to be the left most one..hence it should be 2000000 2)two of the digits have to be 1 and the rest 0..for the number to remain a 7 digit number, the leftmost digit has to be 1..and of the other six positions, 1 of them have to be 1 and the rest 0..there are 6 such possibilities (6C1) Hence answer : 1+6 =7 Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that none of the letters repeat? A. 7 B. 7C7 C. 77 D. 49 E. None of these There are 7! different permutations. Answer : E Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 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 When 10 coins are tossed simultaneously, number of possible outcomes : 2^10 of these half them will have heads in the 3rd coin and half of them will have tail half of 2^10 = 2^10/2= 2^9 Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target? A. 1 B.1/256 C.81/256 D.175/256 E.108/256 IMO, its 1 Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? A.6!/2! B. 3!*3! C.4!/2! D.4!*3!/2! E.3!*3!/2! consider set of vowels as a single unit (vowels)BCS contains 4 units, can be arranged in 4! ways consider the vowels : AAU number of unique arrangements possible : 3!/2! (as there are 2identical elements) Answer : 4!*3!/2! Quote Link to comment Share on other sites More sharing options...
aaadith Posted May 14, 2009 Share Posted May 14, 2009 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 total number of possible ways to fill 6 boxes with either red or green : 2^6=64 all reds not possible. so total possibilities : 63 of these, to satisfy the conditions in the problem, the green balls should be adjacent. no of ways to achieve this when there is 1 green ball : 6 no of ways to achieve this when there are 2 green balls : 5 no of ways to achieve this when there are 3 green balls : 4 no of ways to achieve this when there are 4 green balls : 3 no of ways to achieve this when there are 5 green balls : 2 no of ways to achieve this when there are 6 green balls : 1 total possibilites = sum(1..6)= 6*7/2=21 Quote Link to comment Share on other sites More sharing options...
Ndjamena Posted July 8, 2013 Share Posted July 8, 2013 A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target? A. 1 B.1/256 C.81/256 D.175/256 E.108/256 The probability that he will hit his target on any single shot is 1/4. The probability that he wil NOT hit his target on any single shot is 3/4. The probability that he will hit his target on at least one of the four shots is 1 - P(no hits) = 1 - (3/4)^4 = 1 - 81/256 = 175/256. The correct answer is D. Quote Link to comment Share on other sites More sharing options...
krusta80 Posted March 7, 2014 Share Posted March 7, 2014 (edited) 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 1,000,001 1,000,010 1,000,100 1,001,000 1,010,000 1,100,000 2,000,000 Answer is B In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that none of the letters repeat? A. 7 B. 7C7 C. 77 D. 49 E. None of these The is 7!, which is E (none of these) 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 is half of the outcomes, because half of the time it will be heads. 2^10 / 2 = 2^9 = 512 (E) In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be posted in all of the three post boxes? A. 5 C 3 B. 5 P 3 C. 53 D. 35 E. 25 There are two ways to answer this problem. 1. The easy way... Each letter is independent and has 3 possible destinations (mailboxes). Therefore, there are 3*3*3*3*3 = 3^5 = 243 possible arrangements of five letters into three mailboxes. 2. The hard way... We "count" how many ways to group the letters into three distinct piles and then multiply that by the unique ways to arrange it among the three mailboxes. 0,0,5 --> 1 group * 3 arrangements = 3 ways 0,1,4 --> 5 groups * 6 arrangements = 30 ways 0,2,3 --> 10 groups * 6 arrangements = 60 ways 1,2,2 --> 30 (10*3) groups * 3 (6/2) arrangements = 90 ways 1,1,3 --> 20 (10*2) groups * 3 (6/2) arrangements = 60 ways TOTAL = 243 ways I'm guessing that choice D is supposed to be 3^5 A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target? A. 1 B.1/256 C.81/256 D.175/256 E.108/256 P(miss) = 3/4 P(miss four times) = p = (3/4)^4 = 81/256 P(not missing all four times) = 1-p = 175/256 Answer is D 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 1 Green Ball -> 6 ways 2 Green Balls -> 5 ways 3 Green Balls -> 4 ways 4 Green Balls -> 3 ways 5 Green Balls -> 2 ways 6 Green Balls -> 1 way 21 ways total (B) 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 The wording is a bit off, but I think what they're saying is that spots 1,2, and 4 must NOT be the letter a. If this is the case, then it is 1/4, since there is a 1 in 4 chance that the letter a will be found randomly in spot 3. Answer is A How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R? A. 59 B.11!/3!*2!*2!*2! C. 56 D. 23 E.11!/2!*2!*2! After using an E for the first letter and an R for the last letter, we are left with the following: 1 x m 2 x e 1 x d 1 x i 1 x t 1 x r 2 x a 2 x n For the middle spaces, there are 8*7 = 56 ways permutations of two different letters from the above group, plus 3 ways to repeat them (ee, aa, and nn). Therefore, the total is 59 ways (A) In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? A.6!/2! B. 3!*3! C.4!/2! D.4!*3!/2! E.3!*3!/2! There are 3 distinct ways to group the vowels together (aka in consecutive spaces): AAU AUA UAA There are 4 spaces in which to begin the group of three consecutive vowels: 1 through 4 Since each consonant is distinct, there are 3! = 6 orders possible from left to right (regardless of in which spaces they end up). Finally, we take the product: 3*4*6 = 72 ways (D) Edited March 7, 2014 by krusta80 Quote Link to comment Share on other sites More sharing options...
krusta80 Posted March 7, 2014 Share Posted March 7, 2014 OA-A=7. not sure why Probably because the "!" is missing. Quote Link to comment Share on other sites More sharing options...
krusta80 Posted March 7, 2014 Share Posted March 7, 2014 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 This is a different interpretation than I had for the problem. To me, what you're doing is keeping the order of the consonants the same...NOT the positions. Quote Link to comment Share on other sites More sharing options...
BGPC Eliza Posted October 13, 2014 Share Posted October 13, 2014 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 To elaborate on this further, 10^6 is 1,000,000 and 10^7 is 10,000,000 so we are looking for numbers in between (not including) these two whose digits are equal to 2. There are only 2 options for this: 1) two 1 digits and the rest are 0s, which gives us: 1,000,001 1,000,010 1,000,100 1,001,000 1,010,000 1,100,000 2) one 2 digit and the rest are all 0s, which gives us only 2,000,000 This means that there are a total of 7 numbers (ie B) Quote Link to comment Share on other sites More sharing options...
rajechP Posted October 29, 2014 Share Posted October 29, 2014 In how many ways can the letters of the word “PROBLEM” be rearranged to make 7 letter words such that none of the letters repeat? A. 7 B. 7C7 C. 77 D. 49 E. None of these Hi, The answer is 7P7 = 7!. There may be a typo in option A. There are 7 letters and none is common, so 'n' objects taken all at a time can be arranged (Permutation) in 7P7 ways = 7!. Quote Link to comment Share on other sites More sharing options...
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