In a group 6 girls and 4 boys, four children are to be selected. In how many different ways can they be selected such that at least one girl should be there?

Correct — option (4), 209.

The group has 6 girls and 4 boys, 10 children in all, and 4 are to be selected. The order of choosing does not matter, so this is a count of combinations.

Step 1 — count every selection of 4 from 10.
C(10, 4) = (10 × 9 × 8 × 7) ÷ (4 × 3 × 2 × 1)
= 5,040 ÷ 24 = 210

Step 2 — count the selections with no girl.
No girl means all 4 are boys: C(4, 4) = 1

Step 3 — subtract.
At least one girl = 210 − 1 = 209

Check by cases:
1 girl, 3 boys: C(6, 1) × C(4, 3) = 6 × 4 = 24
2 girls, 2 boys: C(6, 2) × C(4, 2) = 15 × 6 = 90

3 girls, 1 boy: C(6, 3) × C(4, 1) = 20 × 4 = 80
4 girls: C(6, 4) = 15
Total = 24 + 90 + 80 + 15 = 209

The idea to remember: "at least one" = all selections − selections with none.

Why the others are wrong

• (1)
  214 — 214 is more than every possible selection of 4 from the 10 children, which is C(10, 4) = 210.

  A condition such as "at least one girl" can only remove selections from that total, never add to it. The count with the condition is 210 − 1 = 209.

• (2)
  206 — 206 = 210 − 4 would remove four selections. But only one selection has no girl: all four boys, C(4, 4) = 1.

  Subtracting 4 matches C(4, 1), the number of ways to pick one boy, which is not the case being excluded. The case count 24 + 90 + 80 + 15 = 209 confirms that one selection goes.

• (3)
  210 — 210 is C(10, 4), the number of selections with no condition at all.

  It still includes the one group made up of all four boys, which has no girl. Removing that group gives 209.

A combination is a selection in which order does not matter. The number of ways to choose r items from n is C(n, r) = n! ÷ [r! × (n − r)!].

When choices are made from separate groups, multiply: choosing 2 girls from 6 and 2 boys from 4 gives C(6, 2) × C(4, 2) = 90 groups. Separate cases that cannot overlap are added.

"At least one" covers several cases. Counting its opposite, "none", can be a single case, so all selections minus the "none" selections gives the answer in one step.

Useful identity: C(n, r) = C(n, n − r), so C(6, 4) = C(6, 2) = 15.

Key facts

• C(n, r) = n! ÷ [r! × (n − r)!] counts selections of r items from n when order does not matter.
• There are C(10, 4) = 210 ways to select 4 children from 10.
• At least one girl = all selections − selections with no girl = 210 − C(4, 4) = 209.
• Case check for 1, 2, 3 and 4 girls: 24 + 90 + 80 + 15 = 209.
• C(n, r) = C(n, n − r); for example, C(6, 2) = C(6, 4) = 15.

Common traps

• Counting ordered picks. 10 × 9 × 8 × 7 = 5,040 treats the same 4 children chosen in a different order as different groups.
• Forgetting to subtract. C(10, 4) = 210 still includes the all-boys group.
• Fixing one girl first and then choosing any three. C(6, 1) × C(9, 3) = 6 × 84 = 504 counts a group with several girls more than once.

• practice — not a real PYQ

  A committee of 3 is to be formed from 5 men and 3 women so that it has at least one woman. In how many ways can it be formed?

  • (a)56
  • (b)46
  • (c)10
  • (d)30
  Answer(2) — All committees: C(8, 3) = 56. All-men committees: C(5, 3) = 10. At least one woman: 56 − 10 = 46.

  Option (1) ignores the condition, option (3) is the number of all-men committees, and option (4) counts committees with exactly one woman, 3 × C(5, 2) = 30.

• practice — not a real PYQ

  A bag has 4 red and 3 blue balls. Two balls are drawn together at random. What is the probability that at least one of them is red?

  • (a)6/7
  • (b)1/7
  • (c)4/7
  • (d)2/7
  Answer(1) — All pairs: C(7, 2) = 21. Pairs with no red: C(3, 2) = 3. P(at least one red) = 1 − 3/21 = 18/21 = 6/7.

  Option (2) is the probability of no red, option (3) is the chance that a single ball drawn is red, and option (4) is the probability that both are red, C(4, 2)/21 = 6/21.