393]A WORD IS TAKEN TO MEAN ANY ARRANGEMENT OF ONE OR MORE LETTERS.IN HOW MANY WAYS IS IT POSSIBLE TO FORM 6 ENGLISH WORDS,IF THE 6 WORDS TAKEN TOGETHER EACH LETTER IS USED ONE AND ONLYY ONCE.
THIS IS THE THREAD IN WHICH I AM POSTING ALL THOSE MATHS QUESTIONS OF WHICH I COULD NOT GET A CONVINCING SOLUTION BY MYSELF OR OTHERS
AND I REQUEST ALL TARGETIITIANS TO HELP ME OUT
SOME PROBLEMS MIGHT BE TOUGH SOME MIGHT BE VERYYY EASY BUT THESE R THE ONES I COULD NOT SOLVE
SO KINDLY GIVE SOLUTION IN AS MUCH DETAILED WAY AS U CAN
-
UP 0 DOWN 0 2 16
16 Answers
391] HOW MANY BRACELETS STUDDED WITH 18 STUDS CAN BE MADE USING 5 IDENTICAL EMERALDS, 6 IDENTICAL RUBIES, 7 IDENTICAL SAAPHIRES.
392]FIND THE NUMBER OF 6 DIGIT NUMBERS SUCH THAT THE SUM OF THE THREE DIGIT NUMBER FORMED BY THE FIRST THREE DIGITS AND 3-DIGIT NUMBER FORMED BY LAST THREE DIGITS IS LESS THAN 1000.
39
10617!/(2*5!*6!*7!)
1/2 because both clockwise and anticlockwise are same and 17! because it is a circular permutation hence no of permutations is (n-1)!
Here n=18
11Ans for Q2) :
In each of the sets of nos. given below,to get the next no. in the set,add 1 to the no. fomed by the last 3 digits & subtract 1 from the no. formed by the 1st 3 digits.
e.g.In the 1st set,after 999000,comes 998001,and then,997002,...,upto 100899.
999000
to ]---> 900 such nos. (Sum = 999)
100899
998000
to ]---> 899 such nos. (Sum = 998)
100898
997000
to ]---> 898 such nos. (Sum = 997)
100897
.
.
.
.
103000
to ]---> 4 such nos. (Sum = 103)
100003
102000
to ]---> 3 such nos. (Sum = 102)
100002
101000
to ]---> 2 such nos. (Sum = 101)
100001
100000 ]---> 1 such no. (Sum = 100)
∴The total no. of possible 6-digit nos. satisfying the given condition that " THE SUM OF THE THREE DIGIT NUMBER FORMED BY THE FIRST THREE DIGITS AND 3-DIGIT NUMBER FORMED BY LAST THREE DIGITS IS LESS THAN 1000 " is :
900+899+898+...+3+2+1 = (900*901)/2 = 405450 !!
62Question 1] HOW MANY BRACELETS STUDDED WITH 18 STUDS CAN BE MADE USING 5 IDENTICAL EMERALDS, 6 IDENTICAL RUBIES, 7 IDENTICAL SAAPHIRES.
Fix one Sapphire.. then arrange them so that they are not symmetric.. and divide the answer by 2...
add it to the cases when this is symmetric...
Note that it can no longer be identically symmetric until the emerald bead is on the other vertex... (ie the 9th bead)
Not simple at all!! But doable.. how? Does this much hint help?
1Why are you creating threads on your name?
This is ridiculous.
Are you doing it so that your posts are pinked?
33:(
Bad.. this is bad... complain...
No pink worries any one.. this is really bad...
never think of that... i dont think he started this for pink threads... bcause u will get pink if u answers
we are not here for pink posts.. just to help ourselves if not others...
24Yup Abhishek is correct dude..............dont think this way......
62guys dont worry at all.. I am makins sure that no "useless posts" with the intention of being pinked are being pinked...
If there is something really useful that deserves to be pinked... It will be ....
The solutions to all the above posts will be pinked.. soo good luck... :)
39i need 1st question complete soln
its cracking my brain and its not simple formula based that aatmanvor told