37 questions · timed · auto-graded
"b' is the number of permutations of x things taken 11 at a time.
$\therefore\text{b}= \ ^\text{x}\text{P}_{11}$
and, C is the number of permutations of x - 11 things taken all at a time.
$\therefore\ \text{c}= \ ^{ \text{x}-11}\text{P}_{\text{x}-11}$ Now, a = 182bc [Given] $\Rightarrow \ ^{\text{x}+2}\text{P}_{\text{x + 2}} =182\ \times\ ^\text{x}\text{P}_{11} \times\ ^{ \text{x}-11}\text{P}_{\text{x}-11}$ $\Rightarrow (\text{x+2})=182 \times \frac{\text{x!}}{(\text{x}-11)!}\times\text{(x-11)!}\\$ $ \bigg[ \because \ ^\text{n}\text{P}_\text{n}=\text{n!} \text{ and } ^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n-r}!)}\bigg]$ $\Rightarrow (\text{x}+2)! = 182\times\text{x}!$ $\Rightarrow (\text{x}+2)(\text{x}+1)\text{x}! = 182\times\text{x}!$ $\Rightarrow (\text{x}+2)(\text{x}+1) = 182$ $\Rightarrow \text{x}^2+\text{x}+2\text{x}+2 =182$ $\Rightarrow \text{x}^2+3\text{x}+2 -182=0$ $\Rightarrow \text{x}^2+3\text{x}-180=0$ $\Rightarrow \text{x}^2+15\text{x}-12\text{x}-180=0$ $\Rightarrow \text{x}(\text{x}+15)-12(\text{x}+15)=0$ $\Rightarrow (\text{x}-12)(\text{x}-15)=0$ $\Rightarrow \text{x}-12=0\ [\because \text{x}\neq-15]$ $\Rightarrow \text{x}=12$ Hence, $\text{x}=12$Considening 4 vowels as one letter,
We have 7 letters which can be arranged in 7P7 = 7! ways.
A, E, U, I can be put together in 4! ways.
Hence, required number of words = 7! × 4!.
Remaining 5 odd places (1, 3 , 5, 7, 9) are to be occupied by the 6 consonants.
This can be done in 6C5 ways.
Hence, the total number of words in which vowels occupy even places $=\ ^5\text{P}_4 \times \ ^6\text{P}_5$
$=\frac{5!}{(5-4)!}\times \frac{6!}{(6-1)!}$
$=5!\times 6!$
If the words starts with Mand end with I, there are 7 space left for 7 letters.
Number of words that starts with M and end with $\text{I}=\frac{7!}{2!}$
$=7\times5\times4\times3$ $=42\times60$ $=2520$Number of words which do not start with M but end with I
$= 20160 - 2520$ $= 17640$ Required number of words = 17640Hence, total number of signals = 1260.
Number of words starting with H = 5! = 120
Number of words starting with I = 5! = 120
Number of words starting with N = 5! = 120
Number of words starting with T = 5! = 120
Number of words beginning with Z is S!, but one of these words is the word ZEN ITH i tselr. So, we first find the number of words beginning with ZEH, ZEI and ZENH,
Number of words starting with ZEH = 3! = 6
Number of words starting with ZEI = 3! = 6
Number of words starting with ZENH = 2! = 2.
Now, the words beginning with ZENI must follow.
There are 21 words beginning with ZENI one of these words is the word ZENITH itself. The first word beginning with ZENI is the word ZENI HT and the next word is ZENITH.
$\therefore$ Rank of ZENITH= 5 × 120 + 2 × 6 + 2 + 2
= 600 + 12 + 4
= 600 + 16
= 616Numbers
$=\frac{6!}{2!\ 3!}$ $\begin{bmatrix}\because 2\text{ comes} \ \text{2 times and 3 comes 3 times}\end{bmatrix}\\$ $=\frac{6\times5\times4\times3!}{2\times3!}$ $=6\times5\times2$ $=60.$ Now, number of 7-digit numbers $=\frac{7!}{2!\ 3!} = \frac{7\times6\times5\times4\times3!}{2\times3!}$ $=7\times6\times5\times2$ $=420$Hence, total number of numbers which is greater then 1 million= 420 - 60
= 360.
Number of words using 4 letters at a time with no repetitions $=\ ^6\text{P}_4$
$=\frac{6!}{2!}$
$=360$
$=\frac{6!}{(6-6)!}$
$= 6\times 5\times4\times3\times2\times1$
$=720$
$=2\times\ ^5\text{P}_5$
$= 2\times5\times4\times3\times2\times1$
$=240$
There are 6 letters in the word 'VOWELS'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to 6P6 = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720
If we fix up E in the begining then the remaining 5 letters can be arranged in 5P5 = 5! = 5 × 4 × 3 × 2 × 1 = 120 ways
If we fix up O in the begining and Lat the end, the remaining 4 letters can be arranged in 4P4 = 4! = 4 × 3 × 2 × 1 = 24.
There are 2 vowels and 4 consonants in the word 'VOWELS'.
Considering 2 vowels as one letter, we have letters which can be arranged in 5p5 = 5! ways. O, E can be put together in 2! ways.
Hence, required number of words = 5! x 2!
= 5 × 4 × 3 × 2 × 1 × 2 × 1
= 120 × 2
= 240
There are 2 vowels and 4 consonants in the word 'VOWELS'.
Considering 4 consonants as one letter, we have 3 letters which can be arranged in 3P3 = 3! ways.
U, W, L, Scan be put together in 4! ways.
Hence, required number of words in which all consonants come together = 3! x 4!
= 3 × 2 × 4 × 3 × 2
= 144.
$=\frac{8!}{(8-3)!}$
$=\frac{ 6\times 5\times4\times3\times2\times1}{5!}$
$=336$
Hence, the total number three letter words are 336.
P(n, 5) : P(n, 3) = 2 : 1
$\Rightarrow \frac{\text{p}(\text{n},5)}{\text{p}(\text{n},3)}=\frac{2}{1}$ $\Rightarrow\frac{\frac{\text{(n)!}}{(\text{(n-5)!})}}{\frac{\text{n}!}{\text{(n-3)!}}}=\frac{2}{1}$ $\Rightarrow \frac{\text{n}!\times\text{(n-3)!}}{\text{(n-5)!}\times\text{n}!}=2$ $\Rightarrow \frac{\text{(n-3)!}}{\text{(n-5)}!}=2$ $\Rightarrow \frac{(\text{n}-3)(\text{n}-4)(\text{n}-5)!}{(\text{n}-5)!}= 2$ $\Rightarrow (\text{n}-3)(\text{n}-4) =2$ $\Rightarrow \text{n}^2-4\text{n}-3\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12=2$ $\Rightarrow \text{n}^2+7\text{n}+12-2=0$ $\Rightarrow \text{n}^2+7\text{n}+10=0$ $\Rightarrow \text{n}^2-5\text{n}-2\text{n}+10=0$ $\Rightarrow \text{n}(\text{n}-5)-2(\text{n}-5)= 0$ $\Rightarrow (\text{n}-5)(\text{n}-2) =0$ $\Rightarrow \text{n}=5 \ \begin{bmatrix}\ \because\text{r}\ \geq\ 5 \\ \therefore \ \neq\ 2 \end{bmatrix}$ Hence, $\text{n}=5$I = 2 times, T = 2 times, E = 3 times, N, R, M, D, A
Number of letters = 12
The total number of ways when vowels ocuepy even places
$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=\frac{6\times5\times4\times3\times2\times6\times5\times4\times3\times2}{2\times2\times3\times2}$
$=21600$
Required number of ways= 21600
$=\frac{6!}{2!}\times\frac{6!}{2!\ 3!}$
$=21600$
Required number of ways= 21600