Latest Elitmus Aptitude Question SOLUTION: 1,2,3,4,5,6,7 are arranged such that sum of two successive numbers is a prime number....ex-1234765(i.e. 1+2=3, 2+3=5, 3+4=7....) Q1-how many such possible combinations occur? Q-2

Total possible numbers are:
1234765
3214765
5234761
5234167
1674325
5674321
7614325
1476523
3476523
7416523
1256743
3256743
5216743
1652347
5612347
7652143
7652341
1432567
3412567
7432165
7432561
These numbers is obtained with the help of hit & trial in
(_2_4_6_), (_6_4_2_), (_4_6_2_), (_2_6_4_), (_6_2_4_), (_4_2_6_)
Since question uses logic of oeoeoeo.
Q1.) 21
Q2.) 4
Q3.) 3

q2 ans is 4
7432561
7652341
1652347
1432567

I am appreciate to all those who give ideas to treat the problem with regards of odd and even number and their positions. Accordingly the format of the combination will be- "oeoeoeo" where 'o' stands for odd number and 'e' stands for even number.

But, the thing should be notice here is - "2 cun't stay with 7" and also "4 cun't stay with 5" as in both the cases the sum result in 9, which is not a prime number!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Following by my previous post (see above/below) the solution should be such that
a) Total number of arrangement by cosidering the order 'oeoeoeo'
b) Then by removing the terms which occurs as 'succisive 2 and 7' also 'succisive 4 and 5' seperately
c) Then by adding the number of terms those consists of both 'succisive 2 and 7' and 'succisive 4 and 5' because in the last step (b) this types of terms (the intersection) was removed two times instead of one time.

Now executing_ (a)
all the 4 'o' can be permute as 4!
all the 3 'e' can be permute as 3!
hence a = 4!*3!

Now executing_ (b)
first consider the case of 2 followed by 7 {
It can take any of 3 availabe position of 'eo' it can be choosen by 3C1
Rest of_
all the 3 'o' can be permute as 3!
all the 2 'e' can be permute as 2!
hence such types os cases = 3C1*3!*2!
}
now 7 can also follow 2
so, total number of cases in 'succisive 2 and 7' = 2*3C1*3!*2!
similar cases will occus while considering 'succisive 4 and 5'
so, total number of cases under (b)
b = 2*2*3C1*3!*2!

Now exexuting_ (c)
first consider even followed by odd ie. 27 and 45 {
out of 3 'oe' 2 can be filled by 27, 45 and can permute among themselv by 3P2
Rest of_
Both the 2 'o' can be permute as 2!
1 'e' ie. 6 will take the remaining only one vacant position
hence such types os cases = 3P2*2!
}
similar cases will occus when odd is followed by even ie. 72 and 54
hence total number of cases either odd followed by even/ even followed by odd = 2*3P2*2!
now the cases left are_
oeoeoeo
c1_ 72o45eo with 2! ways
c2_ 72oeo45 with 2! ways
c3_ oe72o45 with 2! ways
c4_ o45e72o with 2! ways
and also
c5_ 54o27eo with 2! ways
c6_ 54oeo27 with 2! ways
c7_ oe54o27 with 2! ways
c8_ o27e54o with 2! ways
gives total 8*2! cases. (2! ways because only 2 'o' are able to permute in 2! ways)
so. total number of cases under (c)
c = 2*3P2*2! + 8*2!



Hence result of Q1. is
a-b+c = 4!*3! - 2*2*3C1*3!*2! + (2*3P2*2! + 8*2!)
= 40 numbers of possible combination

I won't repeat the result of Q2, it's correct that answered by Mr. 'AS'
and along with everyone my answer Q3 goes for 3.

Q1.
1674325
5674123
3214765
7614325
5234167
5674321
1234765
Q2. Combinations of 1......7/ 7........1:
0
Q3.
1

see even no. will occupy even position and odd no. will occupu odd position(chk urself)
keeping 4th element 2,4,6 each will give 2 combination and reversing that will give 2 more combination.
(a)4*3=12 combination (m not sure can be more )
(b)2 combinations are posible keeping no.2 at 4th position (1432567 and reversing)
(c)3(as there are 3 even no.)

the 2nd answer is 3 as because the no are 1652347, 7432561,7652341

we...know..e+o=o....,annd here ..total..odd no.=4,and..even ..no.=3..thus..total possible arrengement..willbe.........4*3*3*2*2*1*1=144

@ SANJOY DEBNATH; Excellent approach , but i not understand why are you ignoring the case of (3,6) & 6,3) ?