DAUGHTER Total words =8 ! Total words in which vowels are together =6!×3! words in which all vowels are not together $\begin{aligned}
& =8!-6!\times 3! \
& =6![56-6] \
& =720 \times 50 \
& =36000
\end{aligned}$ Ans.(3)
JEE Main 2025 — Mathematics Algebra
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
Held on 23 Jan 2025 · Verified 6 Jul 2026.
36000
37000
34000
35000
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the set of all values of $k \in \mathbb{R}$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0$, $z \in \mathbb{C}$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:
If $\alpha=1$ and $\beta=1+i\sqrt{2}$, where $i=\sqrt{-1}$ are two roots of the equation $x^3+ax^2+bx+c=0$, $a,b,c \in \mathbb{R}$, then $\int_{-1}^{1}(x^3+ax^2+bx+c)dx$ is equal to:
Let $\alpha = 3+4+8+9+13+14+\ldots$ upto 40 terms. If $(\tan\beta)^{\frac{\alpha}{1020}}$ is a root of the equation $x^2+x-2=0$, $\beta \in \left(0, \dfrac{\pi}{2}\right)$, then $\sin^2\beta + 3\cos^2\beta$ is equal to:
$\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^{2}}{3^{23}}+\ldots+\frac{10 \cdot 2^{24}}{3}$ is equal to :
Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If $P_{n}=C_{0}-C_{1}+\frac{2^{2}}{3} C_{2}-\frac{2^{3}}{4} C_{3}+\ldots. .+\frac{(-2)^{n}}{n+1} C_{n}$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.
Work through every JEE Main Algebra PYQ, year by year.