Question Detail

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

195052
195053
195054
185054

Answer: Option C

Explanation:

\begin{aligned}
^n{P}_r = \frac{n!}{(n-r)!} \\
^{59}{P}_3 = \frac{59!}{(56)!} \\
= \frac{59 * 58 * 57 * 56!}{(56)!} \\
= 195054
\end{aligned}

1. Evaluate combination
\begin{aligned}
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
\end{aligned}

161700
151700
141700
131700

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}

2. Evaluate combination
\begin{aligned}
^{100}{C}_{100}
\end{aligned}

10000
1000
10
1

Answer: Option D

Explanation:

\begin{aligned}
^{n}{C}_{n} = 1 \\
^{100}{C}_{100} = 1
\end{aligned}

3. In how many words can be formed by using all letters of the word BHOPAL

Answer: Option D

Explanation:

Required number
\begin{aligned}
= 6! \\
= 6*5*4*3*2*1 \\
= 720
\end{aligned}

4. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw

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}

5. How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

Answer: Option C

Explanation:

We need to find the ways that vowels NEVER come together.
Vowels are A, E
Let the word be FTR(AE) having 4 words.
Total ways = 4! = 24
Vowels can have total ways 2! = 2
Number of ways having vowel together = 48
Total number of words using all letter = 5! = 120

Number of words having vowels never together = 120-48
= 72

Question Detail

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

1. Evaluate combination \begin{aligned} ^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\ \end{aligned}

2. Evaluate combination \begin{aligned} ^{100}{C}_{100} \end{aligned}

3. In how many words can be formed by using all letters of the word BHOPAL

4. A bag contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the bag, if at least one black ball is to be included in the draw

5. How many words can be formed from the letters of the word "AFTER", so that the vowels never comes together.

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

1. Evaluate combination
\begin{aligned}
^{100}{C}_{97} = \frac{100!}{(97)!(3)!} \\
\end{aligned}

2. Evaluate combination
\begin{aligned}
^{100}{C}_{100}
\end{aligned}