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Miscellaneous Examples · Example 54

Q.From a group of 20 students of an environmental club 9 are to be chosen for an educational tour. There are 3 friends among these students who decide that either all of them will join or none of them will join. In how many ways can be students for the educational tour be chosen?

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The 3 friends must go together or not at all, so the choice splits into two disjoint cases — all 3 friends in (pick remaining 6 from the other 17) or none of the 3 (pick all 9 from the other 17) — whose counts are added.

nCr=n!r!(n−r)!{}^nC_r=\dfrac{n!}{r!(n-r)!}. For an "all-or-none" restriction on a fixed subgroup, split into the two disjoint cases and add their counts (addition principle).

  1. Total students =20=20; the group of 3 friends must be entirely included or entirely excluded; the remaining pool (excluding the 3 friends) has 20−3=1720-3=17 students.

Case A: all 3 friends included

2. The 3 friends occupy 3 of the 9 tour seats, leaving 9−3=69-3=6 seats to fill from the other 17 students: 17C6{}^{17}C_6.

3. 17C6=17×16×15×14×13×126!=8,910,720720=12,376{}^{17}C_6=\dfrac{17\times16\times15\times14\times13\times12}{6!}=\dfrac{8{,}910{,}720}{720}=12{,}376.

Case B: none of the 3 friends included

4. All 9 tour seats are filled from the remaining 17 students: 17C9{}^{17}C_9.

5. 17C9=17C8=17!8! 9!=24,310{}^{17}C_9={}^{17}C_8=\dfrac{17!}{8!\,9!}=24{,}310 (standard binomial value; equivalently built up via Pascal's rule from 17C1=17{}^{17}C_1=17 through 17C8=24,310{}^{17}C_8=24{,}310).

Combine

6. Since the two cases are mutually exclusive and exhaustive, total ways =17C6+17C9=12,376+24,310=36,686={}^{17}C_6+{}^{17}C_9=12{,}376+24{,}310=36{,}686.

✓Final answer

Total ways =17C6+17C9=12,376+24,310=36,686= {}^{17}C_6+{}^{17}C_9 = 12{,}376+24{,}310 = 36{,}686

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