We may have[1 black and 2 non-black] or [2 black and 1 non-black] or [3 black].
| Required number of ways | = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) |
| =[($ 3 \times 5 $ \dfrac{6 \times 5}{2 \times 1} $)]$ + $ [(\dfrac{3 \times 2}{2 \times 1} \times 6]$ + 1 | |
| = (45 + 18 + 1) | |
| = 64. |
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
$\therefore$ Required number of numbers = $\left(1 \times 5 \times 4\right)$ = 20.
The word LEADER contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
| $\therefore$ Required number of ways =$ \dfrac{6!}{(1!)(2!)(1!)(1!)(1!)} $= 360. |
The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.
Given that no digit appears more than once. Hence we have 8 digits remaining (0,1,2,4,6,7,8,9)
So, the next 4 places can be filled with the remaining 8 digits in 8P4 ways.
Total number of ways = 8P4=8×7×6×5=1680
He has 10 patterns of chairs and 8 patterns of tables
A chair can be selected in 10 ways.
A table can be selected in 8 ways.
Hence one chair and one table can be selected in 10×80 ways =80ways
LOGARITHMS contains 10 different letters.
Required number of words= Number of arrangements of 10 letters, taking 4 at a time.
= 10P4=$\left (10 \times 9 \times 8 \times 7\right)$= 5040.
1 red ball can be selected in 4C1 ways.
1 white ball can be selected in 3C1 ways.
1 blue ball can be selected in 2C1 ways.
Total number of ways
= 4C1 × 3C1 × 2C1
=4×3×2=24
He can go in any of the 25 buses [25 ways].
Since he cannot come back in the same bus, he can return in 24 ways.
Total number of ways =25×24=600
| Required number of ways | = (8C5 x 10C6) |
| = (8C3 x 10C4) | |
| = $ \left(\dfrac{8 \times 7 \times 6}{3 \times 2 \times 1} \times\dfrac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} \right) $ | |
| = 11760. |
The first ring can be worn in any of the 3 fingers [3 ways].
Similarly each of the remaining 5 rings also can be worn in 3 ways.
Hence total number of ways
=3×3×3×3×3×3 =$3^6$ =729