Download App

JEE Main 23-Jan-2025 Paper - Shift 1 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 23-Jan-2025 Paper - Shift 14 Marks
MCQ
The number of words, which can be formed using all the letters of the word "DAUGHTER", so that all the vowels never come together, is
  • A
    34000
  • B
    37000
  • ✓
    36000
  • D
    35000
✓

Answer

Correct option: C.
36000
(C)
DAUGHTER
Total words $=8$ !
Total words in which vowels are together $=6!\times 3!$ words in which all vowels are not together$
\begin{array}{l}
=8!-6!\times 3! \\
=6![56-6] \\
=720 \times 50 \\
=36000
\end{array}
$

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 "SECTION - A [MATHS - MCQ]" from JEE Main 23-Jan-2025 Paper - Shift 1→JEE Main 23-Jan-2025 Paper - Shift 1 — all question types→JEE STD 11 Science question papers→Gujarat Board STD 11 Science question papers→Gujarat Board question paper generator→

Similar questions

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non $-$ zero vectors such that $\vec{b}$ and $\vec{c}$ are non $-$ collinear if $\vec{a}+5 \vec{b}$ is collinear with $\overrightarrow{ c }, \overrightarrow{ b }+6 \overrightarrow{ c }$ is collinear with $\overrightarrow{ a }$ and $\overrightarrow{ a }+\alpha \overrightarrow{ b }+\beta \overrightarrow{ c }=\overrightarrow{0}$, then $\alpha+\beta$ is equal to
→
Let $\vec a = 2\hat i + \hat j - 2\hat k$ and $\vec b = \hat i + \hat j$ . Let $\vec c$ be vector such that $\left| {\vec c - \vec a} \right| = 3,\;\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right| = 3$ and the angle between $\vec c$ and $\vec a \times \vec b$ be $30^\circ $ . Then $\vec a \cdot \vec c$ is equal to :
→
Let for $i\, = 1, 2, 3, p_i(x)$ be a polynomial of degree $2$ in $x, p'_i(x)$ and $p"_i(x)$ be the first and second order derivatives of $p_i(x)$ respectively. Let, $A\left( x \right)=\left[ \begin{matrix}
   {{p}_{1}}\left( x \right) & p_{1}^{'}\left( x \right) & p_{1}^{''}\left( x \right)  \\
   {{p}_{2}}\left( x \right) & p_{2}^{'}\left( x \right) & p_{2}^{''}\left( x \right)  \\
   {{p}_{3}}\left( x \right) & p_{3}^{'}\left( x \right) & p_{3}^{''}\left( x \right)  \\
\end{matrix} \right]$ and $B(x)\,= [A(x)]^T$ $A(x)$. Then determinant of $B(x)$
→
The corner points of the bounded feasible region are $(0,0),(2,0),(4,2),(2,4)$ and $\left(0, \frac{10}{3}\right)$

Then for the objective function $z=-x+2 y$

$(i)$ Maximum value of $z$ has at $\ldots \ldots \ldots . . .$

$(ii)$ Minimum value of $z$ has at $\ldots \ldots \ldots . . .$

$(iii)$ The maximum value of $z$ is $\ldots \ldots \ldots . . .$

$(iv)$ The minimum value of $z$ is $\ldots \ldots \ldots . . .$

→
If $A = {\log _2}{\log _2}{\log _4}256 + 2{\log _{\sqrt 2 \,}}\,2,$ then $A$ is equal to
→
The foci of the curve which satisfies the differential equation $(1 + y^2) dx - xy\, dy = 0$ and passes through the point $(1 , 0)$ are :
→
The determinant $\left| {\,\begin{array}{*{20}{c}}{4 + {x^2}}&{ - 6}&{ - 2}\\{ - 6}&{9 + {x^2}}&3\\{ - 2}&3&{1 + {x^2}}\end{array}\,} \right|$ is not divisible by
→
Let $x _1, x _2 \ldots ., x _{100}$ be in an arithmetic progression, with $x _1=2$ and their mean equal to $200$ . If $y_i=i\left(x_i-i\right), 1 \leq i \leq 100$, then the mean of $y _1, y _2$, $y _{100}$ is
→
$\mathop {\lim }\limits_{x \to \infty } \frac{{(2x - 3)(3x - 4)}}{{(4x - 5)(5x - 6)}} = $
→
If $a > 0$ and $z = \frac{{{{\left( {1 + i} \right)}^2}}}{{a - i}}$, has magnitude $\sqrt {\frac{2}{5}} $, then $\bar z$ is equal to:
→