In how many ways can five examination papers be arranged so that … - Vidyadip

MCQ

In how many ways can five examination papers be arranged so that physics and chemistry papers never come together

A
$31$
B
$48$
C
$60$
✓
$72$

✓

Answer

Correct option: D.

$72$

d
(d) Total arrangements of $5$ papers are $ = 5\;! = 120$

When physics and chemistry come together $ = 4\;!\; \times 2\;! = 48$

Hence required number of ways are $120 - 48 = 72$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "MCQ" from permutation and combination→permutation and combination — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $p(n)=x\left(x^{n-1}-n \cdot a^{n-1}+a^n(n-1)\right)$ is divisible by $(x-a)^2$ for:

If the roots of the equation $\frac{1}{{x + p}} + \frac{1}{{x + q}} = \frac{1}{r}$ are equal in magnitude but opposite in sign, then the product of the roots will be

Let $A = \{1, 2, 3\}$. The total number of distinct relations that can be defined over $A$ is

For $A = 133^\circ ,\;2\cos \frac{A}{2}$ is equal to

If $a,\;b,\;c$ are in $G.P.$ and $\log a - \log 2b,\;\log 2b - \log 3c$ and $\log 3c - \log a$ are in $A.P.$, then $a,\;b,\;c$ are the length of the sides of a triangle which is

If ${{3x + a} \over {{x^2} - 3x + 2}} = {A \over {(x - 2)}} - {{10} \over {x - 1}}$, then

Choose the correct answer. If – 3x + 17 < – 13, then:

The number of solutions of the equation $2 \theta-\cos ^{2} \theta+\sqrt{2}=0$ is $R$ is equal to

The remainder when ${5^{99}}$ is divided by $13$ is

If from each of the three boxes containing $3$ white and $1$ black, $2$ white and $2$ black, $1$ white and $3$ black balls, one ball is drawn at random, then the probability that $2$ white and $1$ black ball will be drawn is