The number of words that can be made by re-arranging the letters … - Vidyadip

MCQ

The number of words that can be made by re-arranging the letters of the word APURBA so that vowels and consonants are alternate is:

A
18
B
35
✓
36
D
None of these.

✓

Answer

Correct option: C.

36

The word APURBA is a 6 letter word consisting of 3 vowels that can be arranged in 3 alternate places, in $\frac{3!}{2!}$ ways.
The remaining 3 consonants can be arranged in the remaining 3 places in 3! ways.
$\therefore$ Total number of words that can be formed $=\frac{3!}{2!}\times3!=18$
But this whole arrangement can be set-up in total two ways, i.e either VCVCVC or CVCVCV.
$\therefore$ T otal number of words = 18 × 2 = 36

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 PERMUTATIONS AND COMBINATIONS→PERMUTATIONS AND COMBINATIONS — all question types→MATHS STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

The amplitude of $\frac{1}{\text{i}}$ is equal to:

Choose the correct answer. If $A$ and $B$ are mutually exclusive events, then:

Tangents drawn from origin to the circle ${x^2} + {y^2} - 2ax - 2by + {b^2} = 0$ are perpendicular to each other, if

The orthocentre of the triangle with vertices $\left( {2,\,\frac{{\sqrt 3 - 1}}{2}} \right)$, $\left( {\frac{1}{2},\, - \frac{1}{2}} \right)$ and $\left( {2,\, - \frac{1}{2}} \right)$ is

If $0 \le x < 2\pi $ , then the number of real values of $x,$ which satisfy the equation $\cos x + \cos 2x + \cos 3x + \cos 4x = 0$ is . . .

The number of permutations of n different things taking r at a time when 3 particular things are to be included is:

Choose the correct answer. When tested, the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623 The mean deviations (in hours) from their mean is:

The coefficient of $t^4$ in the expansion of ${\left( {\frac{{1 - {t^6}}}{{1 - t}}} \right)^3}$ is

One mapping is selected at random from all mappings of the set A = {1, 2, 3, ..., n} into itself. The probability that the mapping selected is one to one is:

The number of triplets $(x, y, z)$. where $x, y, z$ are distinct non negative integers satisfying $x+y+z=15$, is