1.Let \[E=\left[\frac{1}{3}+\frac{1}{50}\right]+\left[\frac{1}{3}+\frac{2}{50}\right]+....\] upto 50 terms, then

a) E is divisible by exactly 2 primes

b) E is prime

c) \[E\leq 35\]

d) Both b and c

Explanation:

2. If n < p < 2n and p is prime and \[N=^{2n}C_{n}\] , then

a) \[p\mid N\]

b) \[p^{4}\mid N\]

c) \[p^{2}\mid N\]

d) \[p^{3}\mid N\]

Explanation: n! N = (n + 1) (n + 2) .... (2n)

3. A is a set containing n elements. A subset
\[P_{1}\] of A is chosen. The set A is reconstructed by replacing
the elements of \[P_{1}\] . Next a subset \[P_{2}\] of A is chosen and
again the set is reconstructed by replacing the elements
of \[P_{2}\] . In this way m(> 1) subsets \[P_{1},P_{2},....P_{m}\] of A are
chosen. The number of ways of choosing \[P_{1},P_{2},....P_{m}\] so
that \[P_{i}\cap P_{j}=\phi\] for \[ i\neq j\] , is

a) \[\left(m+1\right)^{n}\]

b) \[2^{m}- ^{m}C_{n}\]

c) \[ \sum_{k=0}^{n}.^{m}C_{k}m^{k}\]

d) Both a and c

Explanation: Let the set A be {a

_{1}, a

_{2}, .... , a

_{n}}. In the first case,

4. With \[P_{1},P_{2},....P_{m}\] as in Q3,
the number of ways of choosing \[P_{1},....P_{m}\] so that
\[P_{1}\cap P_{2}\cap....\cap P_{m}=\phi\] is

a) \[\left(2^{m}-1\right)^{n}\]

b) \[\left(1+2+....+2^{m-1}\right)^{n}\]

c) \[\left(m+1\right)^{n}\]

d) Both a and b

Explanation:

5. The number of 10 digit numbers that can be formed
by using the digits 2 and 3 is

a) \[^{10}C_{2}+^{9}C_{2}\]

b) \[2^{10}\]

c) \[2^{10}-2\]

d) 10!

Explanation: For each place we have two choices

6. If \[P_{r}\] stands for \[^{r}P_{r}\] , then sum of the series \[1+P_{1}+2P_{2}+3P_{3}+....+nP_{n}\] is

a) \[P_{n+1}\]

b) \[P_{n+1}-1\]

c) \[P_{n+1}+1\]

d) none of these

Explanation:

7. If \[a_{n}=\sum_{r=0}^{n}\frac{1}{^{n}C_{r}}\] then value of\[\sum_{r=0}^{n}\frac{n-2r}{^{n}C_{r}}\] is

a) \[\frac{n}{2}a_{n}\]

b) \[\frac{1}{4}a_{n}\]

c) \[na_{n}\]

d) 0

Explanation:

8. m men and w women are to be seated in a row so
that all men sit together. The number of ways in
which they can be seated is

a) (w + 1)! m!

b) m! w!

c) m! (w – 1)!

d) \[^{n}C_{w}\]

Explanation: First treat m men as just one object. We can permute (w + 1) objects in (w + 1)! and the m men is m! ways.

9. A five digit number divisible by 6 is to be formed by using the digits 0, 1, 2, 3, 4 and 8 without repetition.The total number of ways in which this can be done
is

a) 216

b) 150

c) 116

d) 98

Explanation: Use sum of digits must be divisible by 3 and the last digit should be an even number.

10. Rakshit is allowed to select (n + 1) or more books
out of (2n +1) distinct books. If the number of ways
in which he may not select all of them is 63, then
value of n is

a) 3

b) 4

c) 5

d) 11

Explanation: