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
$=1\times8\times7\times2$
$=14\times8=112$
There are 10 persons named $P_1, P_2, P_3, \ldots, P_{10}$. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement $P_1$ must occur whereas $P_4$ and $P_5$ do not occur. Find the number of such possible arrangements.
Given that, $P_1, P_2, \ldots, P_{10}$, are 10 persons, out of which 5 persons are to be arranged but $P_1$ must occur whereas $P_4$ and $P_5$ never occur.
$\therefore$ Selection depends on only $10-3=7$ persons
As, we have already occur $P_1$, Therefore, we have to select only 4 persons out of 7 .
Number of selection $={ }^7 C_4=\frac{7!}{4!(7-4)!}=\frac{7!}{4!3!}=\frac{5040}{24 \times 6}=35$
$\therefore$ Required number of arrangement of 5 persons $=35 \times 5!=35 \times 120=4200$