MATHS M.C.Q (1 Marks)

2. 3720

Solution:

At least one green dye can be chosen in $^5\text{C}_1+\ ^5\text{C}_2\ +\ ^5\text{C}_3+\ ^5\text{C}_4+\ ^5\text{C}_5=22^5-1$ways.

At least one blue dye can be chosen in $^4\text{C}_1+\ ^4\text{C}_2\ +\ ^4\text{C}_3+\ ^4\text{C}_4=2^4-1$ ways.

Any number of red dyes can be chosen in $^3\text{C}_0+\ ^3\text{C}_1\ +\ ^3\text{C}_2+\ ^3\text{C}_3=2^3$ ways.

so, total nnumber of required selection $=(2^5-1)\times(2^4-1)\times2^3=3720$

2. 12

Solution:

Between any two person there is one hand shake.

So, number of hand shakes $=\ ^\text{n}\text{C}_2=\frac{\text{n!}}{2!(\text{n}-2)!}=\frac{\text{n}(\text{n}-1)}{2}=66$(given)

$\Rightarrow\text{n}(\text{n}-1)=132$ $\Rightarrow\text{n}^2-\text{n}-132=0$ $\Rightarrow(\text{n}-12)(\text{n}=11)=0$

$\therefore\text{n}=12$

1. 20

Solution:

Give that $^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_8[\because\ ^\text{n}\text{C}_\text{r}=\ ^\text{n}\text{C}_\text{n-r}]$

$^\text{n}\text{C}_{12}=\ ^\text{n}\text{C}_\text{n-8}$

$\therefore\text{n}-8=12\Rightarrow\text{n}=12+8=20$

3. 7200

Solution:

Given that total number of vowels = 4

and number of consonants = 5

The total of words formed by 2 vowels and 3 consonents

$=\ ^4\text{C}_2\times\ ^5\text{C}_3=\frac{4!}{2!\ 2!}\times\frac{5!}{3!\ 2!}=\frac{4\times3\times2!}{2\times1\times2!}\times\frac{5\times4\times3!}{3!\times2}$

Now permutation of 2 vowels and 3 consonants = 5! = 5 × 4 × 3 × 2 × 1 = 120

So, the total number of words = 60 × 120 = 7200.

2. 144

Solution:

Total number of letters in the ‘ARTICLE’ is 7 out which A, E, I are vowels and R, T, C, L are consonants

Given that vowels occupy even place

$\therefore$ Possible arrangement can be shown as below

C, V, C, V, C, V, C i.e. on 2^nd, 4^th and 6^th places

Therefore, number of arrangement = ³P₃ = 3! = 6 ways

Now consonants can be placed at 1, 3, 5 and 7^th place

$\therefore$ Number of arrangement = ⁴P₄ = 4! = 24

So, the total number of arrangements = 6 × 24 = 144

4. 185

Solution:

Total number of triangle formed from 12 points taking 3 at a time = ¹²C₃

But given that out of 12 points 7 are collinear. So, these seven points will form no triangle.

$\therefore$ The required number of triangles = ¹²C₃ – ⁷C₃

$=\frac{12!}{3!\ 9!}-\frac{7!}{3!\ 4!}=\frac{12\times11\times10\times9}{3\times2\times1\times9!}-\frac{7\times6\times5\times4!}{3\times2\times1\times4!}$

$=\frac{12\times11\times10}{3\times2}-\frac{7\times6\times5}{3\times2}=220-35=185$

3. 24

Solution:

Given digits 2, 3, 4 and 7, we have to form four-digit numbers using these digits.

$\therefore$ Required number of ways = ⁴P₄ = 4! = 4 × 3 × 2 × 1 = 24

3. ¹⁶C₉

Solution:

Total number of players = 22

2 players are always included and 4 are always excluding or never included = 22 – 2 – 4 = 16

$\therefore$ Required number of selection = ¹⁶C₉

2. 64

Solution:

We know that a coin has Head and Tail (H, T)

$\therefore$ When a coin is tossed 6 times, then the

Possible outcome = 2⁶ = 64