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17 Feb 2021, 08:30
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Difficulty:
95%
(hard)
Question Stats:
44% (02:38) correct
56%
(02:24)
wrong
based on 118
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If \((x_1 + x_2 + x_3)(y_1 + y_2) = 77\), what is the number of positive integer solutions of the equation?
(A) 150
(B) 270
(C) 420
(D) 1024
(E) 1048
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
(A) 150
(B) 270
(C) 420
(D) 1024
(E) 1048
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
Number of positive integral solution
satisfying the equation (x1 + x2 + x3) (y1 + y2) =
77, is
(A) 150 (B) 270
(C) 420 (D) 1024
satisfying the equation (x1 + x2 + x3) (y1 + y2) =
77, is
(A) 150 (B) 270
(C) 420 (D) 1024
17 Feb 2021, 16:19
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Bunuel
\((x_1 + x_2 + x_3)(y_1 + y_2) = 77=7*11\)
We are looking for positive integral solutions, meaning none of them will be 0 or negative.
So two cases can be
1) \((x_1 + x_2 + x_3)=7\), that is we have to find ways to distribute 7 things in three parts => (7-1)C(3-1)=6C2=15
\((y_1 + y_2) = 11\), => (11-1)C(2-1)=10C1=10
Total ways = 15*10=150
2) \((x_1 + x_2 + x_3)=11\), that is we have to find ways to distribute 11 things in three parts => (11-1)C(3-1)=10C2=45
\((y_1 + y_2) = 7\), => (7-1)C(2-1)=6C1=6
Total ways = 45*6=270
Total ways = 150+270=420.
You can also find solutions separately for each equation but it will be time consuming
\((y_1 + y_2) = 11\)
Now possibilities are (10,1); (9,2) and so on
Also \((x_1 + x_2+x_3) = 11\)
Here possibilities will be 9-1-1 and ways to fill up 3!/2! and so on.
General Discussion
26 Feb 2024, 07:01
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chetan2u
Can you explain the logic of making it 6C2 instead of 7C3 which is the typical way of distributing 7 items 3 ways
