Evaluate permutation equation
\begin{aligned} ^{59}{P}_3 \end{aligned}

Answer: Option D

Explanation:

\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{75}{P}_2 = \frac{75!}{(75-2)!} \\
= \frac{75*74*73!}{(73)!} \\
= 5550


\end{aligned}

Answer: Option D

Explanation:

Lets suppose there were x number os team in cricket cup then from the given statement we can calculate :
\begin{aligned}
=> ^n C _2 = 153 \\
\text{ it given n = 18 and n = -17} \\
\end{aligned}
As answer cannot be negative to 18 is the answer

Answer: Option A

Explanation:

\begin{aligned}
^{n}{C}_r = \frac{n!}{(r)!(n-r)!} \\
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
= \frac{100*99*98*97!}{(97)!(3)!} \\
= \frac{100*99*98}{3*2*1} \\
= \frac{100*99*98}{3*2*1} \\
= 161700
\end{aligned}

Answer: Option A

Explanation:

From 2 white balls, 3 black balls and 4 red balls, 3 balls are to be selected such that
at least one black ball should be there.

Hence we have 3 choices
All three are black
Two are black and one is non black
One is black and two are non black
Total number of ways
= 3C3 + (3C2 x 6C1) + (3C1 x 6C2) [because 6 are non black]

\begin{aligned}
= 1 + \left[3 \times 6 \right] + \left[3 \times \left(\dfrac{6 \times 5}{2 \times 1}\right) \right]
= 1 + 18 + 45
= 64
\end{aligned}

Answer: Option B

Explanation:

As we can see the letter "E" is twice in given word, so Required Number
\begin{aligned}
= \frac{7!}{2!} \\
= \frac{7*6*5*4*3*2!}{2!} \\
= 2520
\end{aligned}