The number of different words that can be formed from the letters of the word 'INTERMEDIATE' such that two vowels never come together is ........... .
Total number of letters in the word 'INTERMEDIATE' $=12$ out of which 6 are consonants and 6 are vowels. The arrangement of these 12 alphabets in which two vowels never come together can be understand with the help of follow manner.
| V | C | V | C | V | C | V | C | V | C | V | C | V |
6 consonants out of which 2 are alike can be placed in $\frac{6!}{2!}$ ways and 6 vowels, out of which 3 E's alike and 2 I's are alike can be arranged at seven place in ${ }^7 P_6 \times \frac{1}{3!} \times \frac{1}{2!}$ ways.
$\therefore$ Total number of words $=\frac{6!}{2!} \times{ }^7 P_6 \times \frac{1}{3!} \times \frac{1}{2!}=151200$
Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done, if atleast 2 are red, is.
$$\begin{aligned} \text { Required number of ways } & ={ }^5 \mathrm{C}_2 \times{ }^7 \mathrm{C}_1+{ }^5 \mathrm{C}_3 \quad \text{[since, at least two red]}\\ & =10 \times 7+10 \\ & =70+10=80 \end{aligned}$$
The number of six-digit numbers all digits of which are odd, is .......... .
Among the digits $0,1,2,3,4,5,6,7,8,9$, clearly $1,3,5,7$ and 9 are odd.
$\therefore$ Number of six-digit numbers $=5 \times 5 \times 5 \times 5 \times 5 \times 5=5^6$
In a football championship, 153 matches were played. Every two teams played one match with each other. The number of teams, participating in the championship is ............ .
Let the number of team participating in championship be $n$.
Since, it is given that every two teams played one match with each other.
$\therefore$ Total match played $={ }^n C_2$
According to the question,
$$\begin{array}{r} & { }^n C_2 =153 \\ \Rightarrow & \frac{n(n-1)}{2} =153 \\ \Rightarrow & n^2-n =306 \\ \Rightarrow & n^2-n-306 =0 \\ \Rightarrow & (n-18)(n+17) =0 \\ \Rightarrow & n =18,-17 \quad \text{ [inadmissible]}\\ \therefore & n =18 \end{array}$$
The total number of ways in which six ' + ' and four '$-$' signs can be arranged in a line such that no two signs '$-$' occur together, is ............. .
The arrangement can be understand with the help of following figure.
| $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ | $+$ | $-$ |
Thus, '+' sign can be arranged in 1 way because all are identical. and 4 negative signs can be arranged at 7 places in ${ }^7 C_4$ ways.
$$\begin{aligned} \therefore \quad \text { total number of ways } & ={ }^7 C_4 \times 1 \\ & =\frac{7!}{4!3!}=\frac{7 \times 6 \times 5 \times 4!}{3!\times 4!} \\ & =\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways } \end{aligned}$$