Question 7:
Six members of a committee of a school are to be selected from 6 male teachers, 4 female teachers and a male principal. Find the number of different committees that can be formed if
(a) the principal is the chairman of the committee,
(b) there are exactly 2 females in the committee,
(c) there are not more than 4 males in the committee.
Solution:
(a)
If the principal is the chairman of the committee, the remaining number of committee is 5 members.
Hence, the number of different committees that can be formed from the remaining 6 male teachers and 4 female teachers
=10C5=252
(b)
Exactly 2 females in the committee=4C2×7C4=210
(c)
There are not more than 4 males in the committee= 4 males 2 females + 3 males 3 females + 2 males 4 females=7C4×4C2+7C3×4C3+7C2×4C4=210+140+21=371
Six members of a committee of a school are to be selected from 6 male teachers, 4 female teachers and a male principal. Find the number of different committees that can be formed if
(a) the principal is the chairman of the committee,
(b) there are exactly 2 females in the committee,
(c) there are not more than 4 males in the committee.
Solution:
(a)
If the principal is the chairman of the committee, the remaining number of committee is 5 members.
Hence, the number of different committees that can be formed from the remaining 6 male teachers and 4 female teachers
=10C5=252
(b)
Exactly 2 females in the committee=4C2×7C4=210
(c)
There are not more than 4 males in the committee= 4 males 2 females + 3 males 3 females + 2 males 4 females=7C4×4C2+7C3×4C3+7C2×4C4=210+140+21=371
Question 8:
R E A C T
Solution:
(a) Number of arrangements = 5! = 120
(b)
Step 1
If the letters ‘E ’ and ‘A’ have to be placed side by side, they will be considered as one item.
Together with the letters ‘R ’, ‘C ’ and ‘T ’, there are altogether 4 items.
Step 2
The letters ‘E ’ and ‘A ’ can also be arranged among themselves in their group.
Number of arrangements = 2!
Hence, number of arrangements of all the letters of the word ‘REACT’ in which the letters E and A have to be side by side
= 4! × 2!
= 24 × 2
= 48
The diagram below shows five cards of different letters.
(a) Calculate the number of arrangements, in a row, of all the cards.
(b) Calculate the number of these arrangements in which the letters E and A are side by side.
Solution:
(a) Number of arrangements = 5! = 120
(b)
Step 1
If the letters ‘E ’ and ‘A’ have to be placed side by side, they will be considered as one item.
Together with the letters ‘R ’, ‘C ’ and ‘T ’, there are altogether 4 items.
EA R C T
Number of arrangements = 4!Step 2
Number of arrangements = 2!
Hence, number of arrangements of all the letters of the word ‘REACT’ in which the letters E and A have to be side by side
= 4! × 2!
= 24 × 2
= 48