Ncert Solutions Maths class 11th

19. Since no letter is repeated in the word EQUATION.

The permutation of 8 letters taken all at a time

= ⁸P₈

= $\frac{8!}{\left(8-8\right)!}$

= $\frac{8!}{0!}$

= 8! [since, 0! = 1]

= 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1

= 40320

16.The permutation of 8 persons taken 2 positions at a time is

15. The permutation of 5 different digits namely 1, 2, 3, 4, 5 taken 4 at a time is

⁵P₄ = $\frac{5!}{\left(5-4\right)!}$ = $\frac{5!}{1!}$ = 5 * 4 * 3 * 2 * 1 = 120

The permutation of having 2 or 4 at ones place is

²P₁ = $\frac{2!}{\left(2-1\right)!}$ = $\frac{2!}{1!}$ = 1 * 2 = 2

After fixing one of the even number at last digit we can rearrange the remaining four digits taking 3 at a time. i.e.

⁴P₃ = $\frac{4!}{\left(4-3\right)!}$ = $\frac{4!}{1!}$ = 4 * 3 * 2 * 1 = 24

Therefore, total permutation of 4 digit even number using 1, 2, 3, 4, 5

= 24 * 2

= 48

14. The permutation of having even number at the last digit from the given 6 different digits namely 1, 2, 3, 4, 5, 6 to form a 3-digit number is

After taking one of the even number as last digit we can rearrange the remaining 5 digits taking 2 at a time. i.e.

Therefore, The required number = 20 * 3 = 60

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49. Here, out of 100 students, first section has 40 students and the rest I e, 60 students enters in second section.

As me and my friend are among the 100 students.

The no. of ways of selecting 2 students from the 100 students

= 100C₂

(a) When both enters first section if 2 of us are among the 40 students that are to be selected. Similarly, if both enters second section among the 60 students for that section.

(if 2 of us)

Hence, no. of ways of selecting both in same section = 40C₂ + 60C₂

Probability that both of us are in same section

48. Total no. of ticket for lottery sold = 10, 000

 No. of ticket that are awarded prize = 10

So, no. of ticket that are not awarded prize = 10.000 - 10

= 9990

(a) Now, probability of met getting a prize if we buy one ticket

13. For every four-digit number we have to count the permutation of 10 digits namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken 4 at a time

However, these permutation will include those where 0 is at 1000's place.

So, fixing 0 at 1000's place and rearranging the remaining 9 digits taking 3 at a time.

Therefore, The required number = ¹⁰P₄ – ⁹P₃

= 5040 – 504

= 4536 ways

47. Since the die has two faces each mark 1, three faces each marks 2 and one face mark 3.

The possible sample space of outcome is.

S = {1, 2, 3} so, n (s) = 6

(i) P (2) $\frac{3}{6}=\frac{1}{2}$

(ii) P (1 or 3) = P (1) + P (3) = $\frac{2}{6}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

(iii) P (not 3) = 1 P (3) = 1 -$\frac{1}{6}$ = $\frac{6-1}{6}=\frac{5}{6}$