Kumaran1987

If the letter of the word SEQUESTERED are arranged in all possible ways and these words are written out as in a dictionary form, then what is the 50th rank of the word SEQUESTERED is ? Solution: So the word SEQUESTERED can be arranged in dictionary form of letters as D E E E E Q R S S T U Hence the number of words begin with DEEEEQR = 4! / 2! =12 and  the number of words begin with  DEEEEQS = 4! =24 the number of words begin with  DEEEEQT = 4! / 2! =12 12+24+12 =48 so next word U becomes last word hence next words are 49 and 50 will  be DEEEEQURSST is 49th word and DEEEEQURSTS is the 50th word and the answer -->  50th rank of the word SEQUESTERED is " DEEEEQURSTS  "

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If the letter of the word VERMA are arranged in all possible ways and these words are written out as in a dictionary form, then what is the rank of the word VERMA is ? Solution: so the word VERMA letters of V, E, R, M, A In alphabetical order A, E, M, R, V So the number of words begin with A = 4! (leave A and consider remaining letters E, M, R, V) and the number of words begin with  E = 4! (leave A and consider remaining letters A, M, R, V) and the number of words begin with  M  = 4! (leave A and consider remaining letters E, A, R, V) and the number of words begin with  R  = 4! (leave A and consider remaining letters E, M, A, V) after the word begins with V so here after consider two or more words combinations So the number of words begin with VA   = 3! (leave v and A and consider remaining 3 letters E,M, R) VE becomes next character after VA set so below are the dictionary arrangement goes 1) VEAMR 2) VEARM 3) VEMAR 4) VEMRA 5) VERAM finally ...

In how many ways can the letter of the word PROPORTION be arranged by taking 4 letters at a time? Solution: P R O P O R T I O N below I've wrote letters without repeated words and then how many times particular word repeated P R O T I N P R O         O Here (P, P) 2 P, (R, R) 2 R, and (O,O,O) 3 O from the question here we gonna select 4 words hence we have the following chance to select those 4 words Chance 1:   3 O and remaining 1 are different (That means (O,O,O)--> 3 word and (P,R,T,I,N) ---> 1 word from the 5) Chance 2:  two of same words and other two of the  same words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 2 pair of words ) Chance 3: Two of same words and other two are different words (here (O,O,O), (T,T) and (P,P) out of 3 pairs we gonna select 1 pair of word and other 2 word from (R, O, I, N, T) words consider P,P has been selected pair  ) Chance 4: All 4 are different ( here P, R, ...

Microprocess Microconroller mcq 1000+ from Senthil K

root@ubuntu:~# sudo apt-get install mysql-server  Reading package lists... Done Building dependency tree        Reading state information... Done mysql-server is already the newest version. 0 upgraded, 0 newly installed, 0 to remove and 464 not upgraded. The above installation error appearing while I've try to install my sql server, so it shows that installation has been broken, so you need to remove all the my sql server logs which are partially installed on linux, then again install it as the following steps, apt-get purge mysql-server apt-get purge mysql-common rm -rf /var/log/mysql rm -rf /var/log/mysql.* rm -rf /var/lib/mysql rm -rf /etc/mysql # and then: apt-get install mysql-server --fix-missing --fix-broken"