Question

These are 10 points in a plane, out of these 6 are collinear, if $$N$$ is the number of triangles formed by joining these points. then:

A. $$n \leqslant 100$$  
B. $$100 < n \leqslant 140$$
C. $$140 < n \leqslant 190$$
D. $$n > 190$$
Answer :   $$n \leqslant 100$$
Solution :
Number of required triangles
$$\eqalign{ & = \,{\,^{10}}{C_3} - {\,^6}{C_3} \cr & \frac{{10 \times 9 \times 8}}{6} - \frac{{6 \times 5 \times 4}}{6} \cr & = 120 - 20 \cr & = 100 \cr} $$

Releted MCQ Question on
Algebra >> Permutation and Combination

Releted Question 1

$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$     and $$^n{C_{r + 1}} = 126,$$   then $$r$$ is:

A. 1
B. 2
C. 3
D. None of these.
Releted Question 2

Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are

A. 69760
B. 30240
C. 99748
D. none of these
Releted Question 3

The value of the expression $$^{47}{C_4} + \sum\limits_{j = 1}^5 {^{52 - j}{C_3}} $$    is equal to

A. $$^{47}{C_5}$$
B. $$^{52}{C_5}$$
C. $$^{52}{C_4}$$
D. none of these
Releted Question 4

Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is

A. $$^6{C_3} \times {\,^4}{C_2}$$
B. $$^4{P_2} \times {\,^4}{C_3}$$
C. $$^4{C_2} + {\,^4}{P_3}$$
D. none of these

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Permutation and Combination


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