5 Marks Questions

Question 15 Marks

Match each item given under the column $C_1$ to its correct answer given under the column $C_2.$
Using the digits $1, 2, 3, 4, 5, 6, 7,$ a number of $4$ different digits is formed. Find

	$C_1$		$C_2$
$(a)$	how many numbers are formed.	$(i)$	$840$
$(b)$	how many numbers are exactly divisible by $2.$	$(ii)$	$200$
$(c)$	how many numbers are exactly divisible by $25.$	$(iii)$	$360$
$(d)$	how many of these are exactly divisble by $4.$	$(iv)$	$40$

Answer

	$C_1$		$C_2$
$(a)$	how many numbers are formed.	$(i)$	$840$
$(b)$	how many numbers are exactly divisible by $2.$	$(iii)$	$360$
$(c)$	how many numbers are exactly divisible by $25.$	$(iv)$	$40$
$(d)$	how many of these are exactly divisble by $4.$	$(ii)$	$200$

Total of $4$ digit number formed with $ 1, 2, 3, 4, 5, 6, 7 =\ ^7\text{P}_4=\frac{7!}{(7-4)!}=\frac{7\times6\times5\times4\times3!}{3!}=840$
When anumber is divisible by $2=4\times5\times6\times3=360$
Total number which are divisible by $25 = 40$
Total number which are divisible by $4\ ($last two digits is divisible by $4) = 200$

View full question & answer→

Question 25 Marks

Match each item given under the column $C_1$ to its correct answer given under the column $C_2.$
Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:

	$C_1$		$C_2$
$(a)$	Boys and girls alternate.	$(i)$	$5! \times 6!$
$(b)$	No two girls sit together.	$(ii)$	$10! – 5! 6!$
$(c)$	All the girls sit together.	$(iii)$	$(5!)^2+ (5!)^2$
$(d)$	All the girls are never together.	$(iv)$	$2!\ 5!\ 5!$

Answer

	$C_1$		$C_2$
$(a)$	Boys and girls alternate.	$(iii)$	$(5!)^2+ (5!)^2$
$(b)$	No two girls sit together.	$(i)$	$5! \times 6!$
$(c)$	All the girls sit together.	$(iv)$	$2!\ 5!\ 5!$
$(d)$	All the girls are never together.	$(ii)$	$10!\ 5!\ 6!$

Total number of arrangment when boys and girls alternate:$ = (5!)^2 + (5!)^2$

No two girls sit together: $= 5!\ 6!$
All the girls sit never toghether $= 2!\ 5!\ 5!$
All the girls sit never together $= 10!\ 5!\ 6!$

View full question & answer→

Question 35 Marks

Match each item given under the column $C_1$ to its correct answer given under the column $C_2.$
There are $10$ professors and $20$ lecturers out of whom a committee of $2$ professors and $3$ lecturer is to be formed. Find:

	$C_1$		$C_2$
$(a)$	In how many ways committee can be formed.	$(i)$	$^{10}C_2 \times ^{19}C_3$
$(b)$	In how many ways a particular professor is included.	$(ii)$	$^{10}C_2 \times ^{19}C_2$
$(c)$	In how many ways a particular lecturer is included.	$(iii)$	$^9C_1 \times ^{20}C_3$
$(d)$	In how many ways a particular lecturer is excluded.	$(iv)$	$^{10}C_2\times ^{20}C_3$

Answer

	$C_1$		$C_2$
$(a)$	In how many ways committee can be formed.	$(iv)$	$^{10}C_2\times ^{20}C_3$
$(b)$	In how many ways a particular professor is included.	$(iii)$	$^9C_1 \times ^{20}C_3$
$(c)$	In how many ways a particular lecturer is included.	$(ii)$	$^{10}C_2 \times ^{19}C_2$
$(d)$	In how many ways a particular lecturer is excluded.	$(i)$	$^{10}C_2 \times ^{19}C_3$

We have to select $2$ professor out of $10$ and $3$ lecturers out of $20$
$\therefore$ Number of ways of selection $=\ ^{10}C_2 \times ^{20}C_3$
When a paeticular professor is included the number of ways $=\ ^{10 - 1}C_{1}\times ^{20}C_3 =\ ^9C_1 \times ^{20}C_3$
When a particular lecturer is included number of ways $=\ ^{10}C_2 \times ^{19}C_2$
Whan a particular lecturer is excluded, then number of ways $=\ ^{10}C_2 \times ^{19}C_3$

View full question & answer→

Question 45 Marks

Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.

Answer

We have to form 4-digit numbers which are greater than 6000 and less than 7000.
We know that a number is divisible by 5, if at the unit place of the number there is 0 or 5.
So, unit digit can be filled in 2 ways.
The thousandth place can be filled by ‘6’ only.
The hundredth place and tenth place can be filled together in 8 × 7 = 56 ways. So, total number of ways = 56 × 2 = 112

View full question & answer→

Maths 5 Marks Questions

Test yourself on this topic