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.
$$\begin{aligned} \therefore \quad \text { total number of ways } & ={ }^7 C_4 \times 1 \\ & =\frac{7!}{4!3!}=\frac{7 \times 6 \times 5 \times 4!}{3!\times 4!} \\ & =\frac{7 \times 6 \times 5}{3 \times 2 \times 1}=35 \text { ways } \end{aligned}$$
A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box, if atleast one black ball is to be included in the draw is .............. .
Since, there are 2 white, 3 black and 4 red balls. It is given that atleast one black ball is to be included in the draw.
$$\begin{aligned} \therefore \text { Required number of ways } & ={ }^3 \mathrm{C}_1 \times{ }^6 \mathrm{C}_2+{ }^3 \mathrm{C}_2 \times{ }^6 \mathrm{C}_1+{ }^3 \mathrm{C}_3 \\ & =3 \times 15+3 \times 6+1 \\ & =45+18+1=64 \end{aligned}$$
There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is ${ }^{12} C_2-{ }^5 C_2$.
Three letters can be posted in five letter boxes in $3^5$ ways.
In the permutations of $n$ things $r$, taken together, the number of permutations in which $m$ particular things occur together is ${ }^{n-m} P_{r-m} \times{ }^r P_m$.