1
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The number of ways of dividing 15 persons into 3 groups containing 3,5 and 7 persons so that two particular persons are not included into the 5 persons groups is

A

$\frac{117(11!)}{3!(7!)}$

B

${ }^{15} \mathrm{C}_5{ }^{10} \mathrm{C}_3$

C

$90 \times \frac{13!}{7!}$

D

${ }^{15} \mathrm{C}_5{ }^8 \mathrm{C}_3$

2
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

The coefficient of $x^{10}$ in the expansion of $\left(x+\frac{2}{x}-5\right)^{12}$ is

A

1674

B

2132

C

1892

D

862

3
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

Let $S_1=\sum\limits_{j=1}^{10} j(j-1) \cdot{ }^{10} C_j, S_2=\sum\limits_{j=1}^{10} j \cdot{ }^{10} C_j$ and

$$ S_3=\sum\limits_{j=1}^{10} j^2 \cdot{ }^{10} C_j $$

Assertion (A) $S_3=55 \times 2^9$

Reason (R) $S_1=90 \times 2^8$ and $S_2=10 \times 2^8$

A

Both $(A)$ and $(R)$ are true and $R$ is the correct explanation of (A)

B

Both $(A)$ and $(R)$ are true but $(R)$ is not the correct explanation of (A)

C

(A) is true, but (R) is false

D

(A) is false, but (R) is true

4
AP EAPCET 2025 - 26th May Evening Shift
MCQ (Single Correct Answer)
+1
-0

If $\frac{2 x^4-3 x^2+4}{\left(x^2+1\right)\left(x^2+2\right)}=a+\frac{p x+q}{x^2+1}+\frac{m x+n}{x^2+2}$, then $\frac{n}{q}=$

A

$p+m-a$

B

$\frac{p+m}{a}$

C

$\frac{a}{p+m}$

D

$p+m+a$