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a,b,c

by shashank.ism » Tue Feb 09, 2010 12:59 pm
Three elements a, b and c are selected from the set A = 2, 3, 5, 6, 7 to form a three-digit number 'abc', where a < b < c. Similarly, two elements p and q are selected from the set B = 0, 1, 8, 9 to form a two-digit number 'pq', where p > q. Let, M be the total number of all the possible values of 'abc' and N be the total number of all the possible values of 'pq'. What is the value of (M - N)?


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by sanjayism » Wed Feb 10, 2010 3:33 am
SET A=2,3,5,6,7;
>1st assume c is 5
then M1 =3c3=1
>2nd assume c is 6
then M2 = 2c1+1c1 (1st b is 5; 2nd b is 3) =2+1=3
3rd assume c is 7
> then M3=3c1+2c1+1c1=3+2+1=6

M=1+3+6 =10

similarlly, N=1+2+3=6
M-N=10-6 =4

answer 4
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by thephoenix » Wed Feb 10, 2010 5:58 am
shashank.ism wrote:Three elements a, b and c are selected from the set A = 2, 3, 5, 6, 7 to form a three-digit number 'abc', where a < b < c. Similarly, two elements p and q are selected from the set B = 0, 1, 8, 9 to form a two-digit number 'pq', where p > q. Let, M be the total number of all the possible values of 'abc' and N be the total number of all the possible values of 'pq'. What is the value of (M - N)?


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abc can be formed in 5C3 ways and there can be only 1 way to arrange them in dec order so tot 10 ways for M

pq can be formed in 3C1 * 3C1=9 ways

M-N=1

OA pls
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by harsh.champ » Wed Feb 10, 2010 11:17 pm
shashank.ism wrote:Three elements a, b and c are selected from the set A = 2, 3, 5, 6, 7 to form a three-digit number 'abc', where a < b < c. Similarly, two elements p and q are selected from the set B = 0, 1, 8, 9 to form a two-digit number 'pq', where p > q. Let, M be the total number of all the possible values of 'abc' and N be the total number of all the possible values of 'pq'. What is the value of (M - N)?


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Since,the order of the no.s are given(ascending or descending)no permutations can be done and the any unique selection will determine that particular no.
Thus,for abc we have 5C3 = 10
For pq we have 4C2 = 6
[spoiler]Hence, M = 10 - 6 =4 .D is[/spoiler] the ans.
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by harsh.champ » Wed Feb 10, 2010 11:26 pm
thephoenix wrote:
shashank.ism wrote:Three elements a, b and c are selected from the set A = 2, 3, 5, 6, 7 to form a three-digit number 'abc', where a < b < c. Similarly, two elements p and q are selected from the set B = 0, 1, 8, 9 to form a two-digit number 'pq', where p > q. Let, M be the total number of all the possible values of 'abc' and N be the total number of all the possible values of 'pq'. What is the value of (M - N)?


7
6
5
4
1
abc can be formed in 5C3 ways and there can be only 1 way to arrange them in dec order so tot 10 ways for M

pq can be formed in 3C1 * 3C1=9 ways

M-N=1

OA pls
Hey the phoenix ,
I think you made a mistake over there.
abc can be formed in 5C3 ways and there can be only 1 way to arrange them in dec order so tot 10 ways for M
Similarly,we have to consider for pq also.I see you adopted a different aproach for pq whereas the approach for abc(5C3) can also be implemented for pq(4C2).
pq can be formed in 3C1 * 3C1=9 ways
From what I can make out,I think your approach was like this.

Since,we have a zero,it cannot fill the tens digit as we have to construct a 2-digit no.
Hence,tens digit can be filled in 3C1 ways.
Corresponding to it,the units place can be filled in 3C1 ways.
Hence,the answer is 3C1*3C1 ways = 9 ways.
Hence ,M-N=1

But over here,you also took the case when the tens place had 1 and the units place 8.Thus pq being 18.
But,in the ques. it is given that p>q.Over here,1 is not greater than 8.
So,I guess over here you took some extra cases.
Hope this post helps in figuring out the mistake.:)


Also,shashank can you post the OA and the soln. plz!!
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