1.Algebra Booster

5.50 Algebra Booster
fi (n – 9)(n + 8) = 0
fi n = 9, –8
fi n = 9
Hence, the number persons in that room is 9.
17. The number of possible ways
= Co-efficients of x 10 in (x + x 2 + x 3 + … + x 7 ) 4
= Co-efficients of x 6 in (1 + x + x 2 + … + x 6 ) 4
= Co-efficients of
7
6
Ê1
- x ˆ
x inÁ
Ë 1 - x
˜
¯
= Co-efficients of x 6 in (1 – x) –4
= 6–4–1 C 4–1
= 9 C 3
Thus, x = 9 and y = 3.
Hence, the value of
(x – y – 2)
= 9 – 3 – 2
= 4
18. It is given that
n
Â
k = 1
n-
k
3 = 3280
fi (3 n–1 + 3 n–2 + … + 3 + 1) = 3280
fi (1 + 3 + 3 2 + … + 3 n–2 + 3 n–1 ) = 3280
fi
Ê
n
3 - 1ˆ
Á = 3280
Ë 3-
1
˜
¯
fi (3 n – 1) = 3280 ¥ 2
fi 3 n = 6561
fi 3 n = 81 ¥ 81 = 3 4 ¥ 3 4 = 3 8
fi n = 8
Hence, the value of n is 8.
19. The number of possible ways it can be done
Ê 1 1 1 1ˆ
= 4! Á1 - + - +
Ë 1! 2! 3! 4! ˜
¯
Ê 1 1 1ˆ
= 4! Á - +
Ë 2! 3! 4! ˜
¯
Ê1 1 1 ˆ
= 24 ¥ Á - +
Ë
˜
2 6 24¯
Ê12 - 4 + 1ˆ
= 24 ¥Á
Ë
˜
24 ¯
= 9
20. Clearly, 6 n – 3 n = 189
fi 6 n – 3 n = (216 –27)
fi 6 n – 3 n = 6 3 –3 3
fi n = 3
Hence, the value of n is 3.
4
Previous Years’ JEE-Advanced Examinations
1. Method R 1
R 2
R 3
Number of ways
I 1 3 2 2
C 1
¥ 4 C 3
¥ 2 C 2
= 8
II 1 4 1 2
C 1
¥ 4 C 4
¥ 2 C 1
= 4
III 2 2 2 2
C 2
¥ 4 C 2
¥ 2 C 2
= 6
IV 2 3 1 2
C 2
¥ 4 C 3
¥ 2 C 1
= 8
Thus, the total possible ways
= 8 + 4 + 6 + 6
= 26
2. Each player will get 13 cards.
The number of ways of distributing 52 cards giving 13
cards to each player
= 52 C 13
¥ 39 C 13
¥ 26 C 13
¥ 13 C 13
(52)!
=
4
{(13)!}
3. The number of possible ways of dividing cards into
four groups
52 35 18
C17 C17 C17
¥ ¥ ¥ 1
=
3!
(52)!
=
3
3! ¥ {(17)!}
4. Number of five letter words that can be formed from
the 10 letters = 10 ¥ 10 ¥ 10 ¥ 10 ¥ 10 = 10 5
Number of 5 letter words that have none of their letter
repeated = 10 P 5
= 30240
Thus, the number of words which have at least one of
their letter repeated = 10 5 – 30240 = 69760
5. Clearly B and C have the same number of elements.
6.
5
47 52-
j
C4+ Â C3
j = 1
= 47 C 4
+ 51 C 3
+ 50 C 3
+ 49 C 3
+ 48 C 3
+ 47 C 3
= ( 47 C 4
+ 47 C 3
) + 51 C 3
+ 50 C 3
+ 49 C 3
+ 48 C 3
= ( 48 C 4
+ 48 C 3
) + 51 C 3
+ 50 C 3
+ 49 C 3
= ( 49 C 4
+ 49 C 3
) + 51 C 3
+ 50 C 3
= ( 50 C 4
+ 50 C 3
) + 51 C 3
= ( 51 C 4
+ 51 C 3
)
= 52 C 4
7. The total number of possible ways
= Total number of onto functions between two
sets consists of 5 and 3 elements respectively
= total number of possible ways to distribute
5 balls into 3 boxes, where no box is remain
empty.
5! 3! 5! 3!
= ¥ + ¥
1!1!3! 2! 1! 2! 2! 2!
120 120 ¥ 6
= +
2 8
= 60 + 90
= 150