india_chintan Posted April 24, 2009 Share Posted April 24, 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 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 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 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 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 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 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 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 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! 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! Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 2009 Q. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? A.6!/2! C.4!/2! D.4!*3!/2! B. 3!*3! E.3!*3!/2! Answer is 4!* 3!/2! Explanation: If we take all vowel as a single letter then BCS(AAU) total 4 letter can be arranged in 4! way. Now three vowel can be arranged in 3!/2! ways. ( As in three vowel there is 2 'A' for this it is divided by 2!) So total way is 4!*3!/2! Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 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 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 Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 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 Answer-B=7 Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 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 Answer should be E. Because the actual result should 7! Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 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 Answer is B=21 (6+5+4+3+2+1) Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 24, 2009 Share Posted April 24, 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 The answer should 29. Is answer B 29 or 29???? Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 2009 Q. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together? A.6!/2! C.4!/2! D.4!*3!/2! B. 3!*3! E.3!*3!/2! Answer is 4!* 3!/2! understood the 4!*3! i.e. 3! is for vowels and 4! is for all letters. y is it divided by 2!. Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 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 Answer is 1/4 Total arrangement=4! With consonant in the same order=4 So probability is 4!/4 plz explain. i thought the answer is 1/24. Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 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 Answer should be E. Because the actual result should 7! OA-A=7. not sure why Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 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 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. Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 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 Answer is B=21 (6+5+4+3+2+1) explanation plz?? Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 24, 2009 Author Share Posted April 24, 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 The answer should 29. Is answer B 29 or 29???? spot on buddy. oa-b. but plz post the explanation. thx in advance. Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 26, 2009 Share Posted April 26, 2009 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 Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 26, 2009 Share Posted April 26, 2009 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 Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 26, 2009 Share Posted April 26, 2009 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 Quote Link to comment Share on other sites More sharing options...
india_chintan Posted April 27, 2009 Author Share Posted April 27, 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 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. ?? Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 27, 2009 Share Posted April 27, 2009 u said is answer is 1/4 nd caluclated it as 1/6. ?? I made mistake to write actual answer should be 1/6. Quote Link to comment Share on other sites More sharing options...
lazyjava Posted April 28, 2009 Share Posted April 28, 2009 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 Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted April 29, 2009 Share Posted April 29, 2009 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 Quote Link to comment Share on other sites More sharing options...
givinggmat Posted May 8, 2009 Share Posted May 8, 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 Answer should be E. Because the actual result should 7! The question says - 'none of the letters repeat'...this shuld be reg the positions(atleast from OG perspective). so, according to OG it is 7 instead of 7! 1 -- PROBLEM 2 -- ROBLEM P 3 -- 0BLEM PR and so on.... which will stop after 7th combination. The Answer - A Not sure how excatly do we need to interpret this question...!!! Any inputs...? thank you. Quote Link to comment Share on other sites More sharing options...
Md. Minuddin Posted May 8, 2009 Share Posted May 8, 2009 The question says - 'none of the letters repeat'...this shuld be reg the positions(atleast from OG perspective). so, according to OG it is 7 instead of 7! 1 -- PROBLEM 2 -- ROBLEM P 3 -- 0BLEM PR and so on.... which will stop after 7th combination. The Answer - A Not sure how excatly do we need to interpret this question...!!! Any inputs...? thank you. PROBLEM has 7 letter which is different from each other 1----2---3----4---5----6---7 Actually I think repeat means using same letter more then once. If so then for first position we have 7 choice. As we can't repeat same letter so now from the rest 6 letter we can choice second position in 6 way And so on, So total way 7*6*5*4*3*2*1=7! 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 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 Quote Link to comment Share on other sites More sharing options...
abhishek_mumbai 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 Answer should be 2^9.. Quote Link to comment Share on other sites More sharing options...
abhishek_mumbai 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 1/4 * 1/4 * 1/4 * 1/4 = 1/256 Quote Link to comment Share on other sites More sharing options...
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