1Maths_Psy....it's 1536 ....pls post ur method
1) How many arrangements can be made with all the letters of the word "VENUS" such that the order of the vowels remains unaltered ?
2) In how many ways can 5 commerce and 4 science students be arranged in a row so that the commerce and the science students are placed alternately ?
3)How many nos. greater than 1,00,000 can be formed with the digits 0,2,5,2,4,5.
4) How many 6-digit even nos. can be formed from 2,3,5,3,4,5 alone ?
5) Don't solve this ; just explain the meaning :
A code-signal uses digit-letter-digit combinations but 0 and 1 are not used for either digits or letters. How many code signals are possible ?
6)Again this one : just explain the meaning :
In how many ways can the letters of the word "CONTACT" be arranged
(i) without changing the order of the vowels
(ii) without changing the relative positions of vowels and consonants
(iii) without changing the positions of the vowels
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34 Answers
1b)it means consonants at 1st,3rd,4rth,6th n 7th positions n vowels at remaining
1for 2nd the no of commerce students are more than no of science students'
so the commerce student ll be ist one in row '
so u ve to just write the no of ways in which science n cmmrce students can be shuffled among themselves
thatis 5!(for ccmmmrce)*4!(for science)
1is d answer to d 5th question 1664 or is it 1536...if ne1 is correctyes,den i can explain d solution...actually d question is a bit difficult to understand...
1Eurie....I too thought upon this logic....but i was not sure....thnx for ur explanation.
1c aveek......its told to omit '0' and '1' in the digits place,but in case of alphabets u also have to omit 'o' and 'i'....so answer is 8*24*8.....now did u get it??......
1i look at this problem like this.......
VENUS.......total 5 letters........total ways of arranging them=5P5.....
but in this case E can come in front of U or U can come before E......so to nullify the 2nd case we divide it by 2! ........so ans is 5P5/2!.......
1i look into the problem as thus....
_S_S_S_S_..........placing the science students first.......then there are 5 probable alternate places where the commerce students can be seated......the students can be permutated in 5P5 no. of ways......similarly the science stdnts can be permutated in 3P3 no of ways for each arrangement of sc stdnts......so the total no of possible arrangements is (3P3X5P5).....
if we had placed the commerce stdnts first then _C_C_C_C_C_............a case could hav arised that S C S C S C S C C........here 2 commerce stdnts are placed together which is wrong according to the question.....so instead the science stdnts are placed first.....
62subhomoy... eureka said the exactly same thing :)
btw I could not make any head or tail of the 5th part either..
1) How many arrangements can be made with all the letters of the word "VENUS" such that the order of the vowels remains unaltered ?
3)How many nos. greater than 1,00,000 can be formed with the digits 0,2,5,2,4,5.
Isnt the answer infinite?
4) How many 6-digit even nos. can be formed from 2,3,5,3,4,5 alone ?
5!/(2x2)
5) Don't solve this ; just explain the meaning :
A code-signal uses digit-letter-digit combinations but 0 and 1 are not used for either digits or letters. How many code signals are possible ?
No Head or tail...
49nishant bhaiya.............u asked this question?????????????but it means...
there are 3 dashes......D L D
the one named D can be filled by digits.........except 0 or 1 i.e. 2,3,4,5,6,7,8,9...........the one named L can be filled by 26 english alphabets.....so wheres the problem so why wait?????? solve it.....
I am hiding something don't see until u have solved it...
8*26*8 gives the answer......8 nos * 26 letters * 8 nos again
u kno wat.....it feels great to revive a thread..... n i love PER-COM...
24this is detail method
case I :
when E occupies the first position, no. of words = 4! = 24
case II :
when E occupies 2nd pos., no of words = 3*3*2*1= 18
case III :
when E occupies the 3rd pos. , no of words = 3*2*2*1 = 12
case IV :
when E occupies the 4th position, no of words =3! = 6
So, total no. of words possible = 24 + 18 + 12 + 6 = 60
if u want shorter method ..tehn i can give that
1out of 5 places choose three places for consonenets
5c3*3!(ways to arrange))
1for 3rd i m getting 150
for 4rth i got 60
i m in hurry so may be calc mistake
1150 and 60 are correct. Pls show the method ....pleeeeeeeeeeeease
14 total no of digits ending with 2+nno of digits ending with 4
5!/(2!*2!)+5!/(2!*2!)