samarpan.g28 wrote:
Purnank wrote:
GMATWhizTeam wrote:
a + b + c = 34, where a, b and c are primes. How many ordered solution sets of (a, b, c) are possible, for the given equation?
(A) 0
(B) 2
(C) 3
(D) 6
(E)12
(A) 0
(B) 2
(C) 3
(D) 6
(E)12
a + b + c = 34 (even) only possible when even + odd + odd = even.
therefore the even prime that we have is 2 and now equation will become,
odd + odd = 34 - 2 = 32.
set of possible values of those two odd primes (3, 29) and (13, 19) from the set of (3,5,7,11,13,17,19,23,29)
hence possible set of values (2,3,29) and (2,13,19)
and their arrangements for each possibility is 3!.
Therefore total possibilities are 2*3! = 12.
HenceE.
Hi. If it is an ordered set, can we consider 3! for each set for theirarrangements?
Yes 3! for each
like for 2 3 29
can be arranged in (2 3 29) (2 29 3) (3 29 2) (3 2 29) (29 2 3) (29 3 2)
similary for 2 13 19 you can make 6 possible arrangements.
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Statistics : Posted by Purnank • on 08 Oct 2021, 10:26 • Replies 7 • Views 2275
