How many automobile license plates can be made, if each plate contains two different letters followed by three different digits?
There are 26 English alphabets and 10 digits (0 to 9).
Since, it is given that each plate contains two different letters followed by three different digits.
$\therefore$ Arrangement of 26 letters, taken 2 at a time $={ }^{26} P_2=\frac{26!}{24!}=26 \times 25=650$
and three-digit number can be formed out of the 10 digits $={ }^{10} P_3=10 \times 9 \times 8=720$ ways
$\therefore$ Total number of licence plates $=650 \times 720=468000$
A bag contains 5 black and 6 red balls, determine the number of ways in which 2 black and 3 red balls can be selected from the lot.
It is given that bag contains 5 black and 6 red balls.
So, 2 black balls is selected from 5 black balls in ${ }^5 \mathrm{C}_2$ ways.
and 3 red balls are selected from 6 red balls in ${ }^6 C_3$ ways.
$\therefore$ Total number of ways in which 2 black and 3 red balls are selected $={ }^5 C_2 \times{ }^6 C_3$
$$=10 \times 20=200 \text { ways }$$
Find the number of permutations of $n$ distinct things taken $r$ together, in which 3 particular things must occur together.
Total number of things $=n$
We have to arrange $r$ things out of $n$ in which three things must occur together.
Therefore, combination of $n$ things taken $r$ at a time in which 3 things always occurs
$$={ }^{n-{ }^3} C_{r-3}$$
If three things taken together, then it is considered as 1 group.
Arrangement of these three things $=3$ !
Now, we have to arrange $=r-3+1=(r-2)$ objects
$\therefore \quad$ Arranged of $(r-2)$ objects $=r-2$ !
$\therefore$ Total number of arrangements $={ }^{n-3} C_{r-3} \times r-2!\times 3!$
Find the number of different words that can be formed from the letters of the word 'TRIANGLE', so that no vowels are together.
Number of letters in the word 'TRIANGLE' $=8$, out of which 5 are consonants and 3 are vowels.
If vowels are not together, then we have following arrangement.
| V | C | V | C | V | C | V | C | V | C | V |
Consonants can be arranged in $=5!=120$ ways and vowels can occupy at 6 places.
The 3 vowels can be arranged at 6 place in ${ }^6 P_3$ ways $=\frac{6!}{6-3!}=\frac{6!}{3!}$
$$=\frac{6 \times 5 \times 4 \times 3!}{3!}=120$$
Total number of arrangement $=120 \times 120=14400$
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5 , provided that no digit is to be repeated.
We know that a number is divisible by 5 , If at the units place of the number is 0 or 5 . We have to form 4 -digit number which is greater than 6000 and less than 7000 . So, unit digit can be filled in 2 ways.
Since, repeatition is not allowed. Therefore, tens place can be filled in 7 ways, similarily hundreds place can be filled in 8 ways.
But we have to form a number greater than 6000 and less than 7000 .
Hence, thousand place can be filled in only 1 ways.
| 6 | 8 | 7 | 2 |
Total number of integers
$=1\times8\times7\times2$
$=14\times8=112$