MCQ
The sum $\sum\limits_{k=1}^{20}(1+2+3+\ldots+k)$ is
- A$1496$
- B$1690$
- ✓$1540$
- D$1560$
$=\frac{1}{6} \times 20 \times 21 \times 22=1540$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(i)$ Let $\alpha_1$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, having exactly $3$ boys and $2$ girls.
$(ii)$ Let $\alpha_2$ be the total number of ways in which the committee can be formed such that the committee has at least $2$ members, and having an equal number of boys and girls.
$(iii)$ Let $\alpha_3$ be the total number of ways in which the committee can be formed such that the committee has $5$ members, at least $2$ of them being girls.
$(iv)$ Let $\alpha_4$ be the total number of ways in which the committee can be formed such that the committee has $4$ members, having at least $2$ girls and such that both $M _1$ and $G _1$ are $NOT$ in the committee together.
| $LIST I $ | $LIST I $ |
| $P$ The value of $\alpha_1$ is | $1$ $136$ |
| $Q$ The value of $\alpha_2$ is | $2$ $189$ |
| $R$ The value of $\alpha_3$ is | $3$ $192$ |
| $S$ The value of $\alpha_4$ is | $4$ $200$ |
| $5$ $381$ | |
| $6$ $461$ |
The correct option is: